Comparing the motions of two bodies

One formula or one step of reasoning. Core exercises first, the rest is extra practice.

1. Exercise 3-1Core

   Two carts move on the same straight track. The position of cart A is $x_\text{A} = 4.0\,\text{m} + 2.0\,\text{m/s} \cdot t$ and the position of cart B is $x_\text{B} = 24\,\text{m} - 3.0\,\text{m/s} \cdot t$. When do the carts meet?

   sCheck
   Answer

   $4.0\,\text{s}$

   Show the solution

   Given

   $x_\text{A} = 4.0\,\text{m} + 2.0\,\text{m/s} \cdot t$ and $x_\text{B} = 24\,\text{m} - 3.0\,\text{m/s} \cdot t$ in the same coordinate system

   Find

   the time of the meeting $t$ in $\text{s}$

   Principle

   The carts meet when they are at the same place at the same moment, so $x_\text{A} = x_\text{B}$. The positions are linear functions of time, so the equation has one solution.

   Formula

   $x_\text{A} = x_\text{B}$, so $4.0 + 2.0\,t = 24 - 3.0\,t$. Terms in $t$ to the left and numbers to the right: $5.0\,t = 20$, so $t = \dfrac{24 - 4.0}{2.0 + 3.0}$

   Conversions

   The positions are in metres and the velocities in $\text{m/s}$, so the result comes out in seconds. No conversions are needed.

   Substitution

   $t$$= \dfrac{24\,\text{m} - 4.0\,\text{m}}{2.0\,\text{m/s} + 3.0\,\text{m/s}}$

   $= \dfrac{20\,\text{m}}{5.0\,\text{m/s}}$

   $= 4.0\,\text{s}$

   Answer

   $4.0\,\text{s}$

   Two significant figures like the velocities ($2.0\,\text{m/s}$ and $3.0\,\text{m/s}$).

   Reasonableness

   Substituting into both positions:

   $x_\text{A}$$= 4.0\,\text{m} + 2.0\,\text{m/s} \cdot 4.0\,\text{s}$

   $= 12\,\text{m}$

   $x_\text{B}$$= 24\,\text{m} - 3.0\,\text{m/s} \cdot 4.0\,\text{s}$

   $= 12\,\text{m}$

   The positions are equal, so the carts meet. The time is positive, so the meeting happens after the start.

2. Exercise 3-2Core

   A cyclist rides from the place $0$ at a constant velocity of $8.0\,\text{m/s}$ in the positive direction. At the same moment a runner is at the place $60\,\text{m}$ and runs in the same direction at a constant velocity of $5.0\,\text{m/s}$.

   When does the cyclist catch up with the runner?

   sCheck
   Answer

   $20\,\text{s}$

   Show the solution

   Given

   cyclist $x_0 = 0$, $v_\text{C} = 8.0\,\text{m/s}$; runner $x_0 = 60\,\text{m}$, $v_\text{R} = 5.0\,\text{m/s}$; both uniform and in the same direction

   Find

   the time $t$ when the cyclist catches up with the runner, in $\text{s}$

   Principle

   Catching up is a meeting: the positions are equal. Both move uniformly, so $x = x_0 + v\,t$ for each. The cyclist is faster, so she catches up with the runner.

   Formula

   $v_\text{C}\,t = x_{0\text{R}} + v_\text{R}\,t$, so

   $(v_\text{C} - v_\text{R})\,t$$= x_{0\text{R}}$

   $t$$= \dfrac{x_{0\text{R}}}{v_\text{C} - v_\text{R}}$

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   $t$$= \dfrac{60\,\text{m}}{8.0\,\text{m/s} - 5.0\,\text{m/s}}$

   $= \dfrac{60\,\text{m}}{3.0\,\text{m/s}}$

   $= 20\,\text{s}$

   Answer

   $20\,\text{s}$

   Two significant figures like the velocities.

   Reasonableness

   The cyclist gains $3.0\,\text{m}$ on the runner every second, so $60\,\text{m}$ takes $20\,\text{s}$. Check: the position of the cyclist is $8.0\,\text{m/s} \cdot 20\,\text{s} = 160\,\text{m}$ and that of the runner is $60\,\text{m} + 5.0\,\text{m/s} \cdot 20\,\text{s} = 160\,\text{m}$.

3. Exercise 3-3

   A car starts from rest at the origin and accelerates uniformly at $3.0\,\text{m/s}^2$. At what position is the car after $4.0\,\text{s}$?

   mCheck
   Answer

   $24\,\text{m}$

   Show the solution

   Given

   $x_0 = 0$, $v_0 = 0$, $a = 3.0\,\text{m/s}^2$, $t = 4.0\,\text{s}$

   Find

   the position $x$ of the car in $\text{m}$

   Principle

   The acceleration is constant, so the position comes from the position function of uniformly accelerated motion. The car starts from rest at the origin, so the initial position and the initial velocity are zero.

   Formula

   $x = x_0 + v_0\,t + \tfrac12\,a\,t^2$, and when $x_0 = 0$ and $v_0 = 0$, $x = \tfrac12\,a\,t^2$

   Conversions

   All quantities are in SI units, and $\text{m/s}^2 \cdot \text{s}^2 = \text{m}$. No conversions are needed.

   Substitution

   $x$$= \tfrac12 \cdot 3.0\,\text{m/s}^2 \cdot (4.0\,\text{s})^2$

   $= \tfrac12 \cdot 3.0\,\text{m/s}^2 \cdot 16\,\text{s}^2$

   $= 24\,\text{m}$

   Answer

   $24\,\text{m}$

   Two significant figures like the given values.

   Reasonableness

   If the car moved all the time at its final velocity $3.0\,\text{m/s}^2 \cdot 4.0\,\text{s} = 12\,\text{m/s}$, it would advance $48\,\text{m}$. In accelerated motion the average velocity is half of the final velocity, so the distance is half of that, $24\,\text{m}$.

4. Exercise 3-4

   The graph shows the positions of two carts as functions of time. Read from the graph when the carts meet.

   sCheck
   Answer

   $5.0\,\text{s}$

   Show the solution

   Given

   a graph of the positions of the carts: cart A starts from $0$ and cart B from $40\,\text{m}$

   Find

   the time of the meeting $t$ in $\text{s}$

   Principle

   The carts are at the same place at the same moment at the point where the lines of the $(t,x)$ graph intersect. The time of the meeting is the horizontal coordinate of the intersection.

   Reading the graph

   The lines intersect at the point at the horizontal grid line $x = 30\,\text{m}$. From there a vertical line is dropped to the time axis: it hits $t = 5.0\,\text{s}$, halfway between two grid lines ($4$ and $6\,\text{s}$).

   Formula

   $t = $ the horizontal coordinate of the intersection, read from the graph

   Conversions

   The axes are already in $\text{s}$ and $\text{m}$, so no conversions are needed.

   Substitution

   The pair read is $(t$; $x) = (5.0\,\text{s}$; $30\,\text{m})$. A calculation check from the end points of the graph: A advances $60\,\text{m}$ in $10\,\text{s}$, so $v_\text{A} = 6.0\,\text{m/s}$; B moves from $40\,\text{m}$ to $20\,\text{m}$, that is, $-20\,\text{m}$ in $10\,\text{s}$, so $v_\text{B} = -2.0\,\text{m/s}$. The equation $6.0\,t = 40 - 2.0\,t$ gives

   $t$$= \dfrac{40\,\text{m}}{8.0\,\text{m/s}}$

   $= 5.0\,\text{s}$

   Answer

   $5.0\,\text{s}$

   Read from the graph to two significant figures.

   Reasonableness

   The place of the meeting

   $x$$= 6.0\,\text{m/s} \cdot 5.0\,\text{s}$

   $= 30\,\text{m}$

   is between the two starting places ($0$ and $40\,\text{m}$), as it should be. Cart A has travelled $30\,\text{m}$ and B $10\,\text{m}$, and A is three times as fast.

5. Exercise 3-5

   Runner A starts from the place $0$ and runs at a constant velocity of $3.5\,\text{m/s}$ in the positive direction. Runner B is at the same moment at the place $60\,\text{m}$ and runs toward A at a constant velocity of $-2.5\,\text{m/s}$.

   At what place do they meet?

   mCheck
   Answer

   $35\,\text{m}$

   Show the solution

   Given

   runner A: $x_{0\text{A}} = 0$, $v_\text{A} = 3.5\,\text{m/s}$; runner B: $x_{0\text{B}} = 60\,\text{m}$, $v_\text{B} = -2.5\,\text{m/s}$; both move uniformly

   Find

   the place of the meeting $x$ in $\text{m}$

   Principle

   The runners meet when they are at the same place at the same moment, so $x_\text{A} = x_\text{B}$. Once the time of the meeting is solved, the place of the meeting is the position of either runner at that moment. The velocity of B is negative, because B runs toward A, that is, in the negative direction.

   Formula

   $v_\text{A}\,t = x_{0\text{B}} + v_\text{B}\,t$, so

   $t$$= \dfrac{x_{0\text{B}}}{v_\text{A} - v_\text{B}}$

   and the place of the meeting is $x = v_\text{A}\,t$

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   $t$$= \dfrac{60\,\text{m}}{3.5\,\text{m/s} - (-2.5\,\text{m/s})}$

   $= \dfrac{60\,\text{m}}{6.0\,\text{m/s}}$

   $= 10\,\text{s}$

   $x$$= 3.5\,\text{m/s} \cdot 10\,\text{s}$

   $= 35\,\text{m}$

   Answer

   $35\,\text{m}$

   Two significant figures like the given values.

   Reasonableness

   The place is between the starting places of the runners, $0$ and $60\,\text{m}$, as it should be. The position of B at the same moment: $60\,\text{m} - 2.5\,\text{m/s} \cdot 10\,\text{s} = 35\,\text{m}$. A has run $35\,\text{m}$ and B $25\,\text{m}$, together $60\,\text{m}$.

6. Exercise 3-6

   The $(t,v)$ graphs describing the velocities of two cars are drawn in the same coordinate system. The cars drive along the same straight road in the same direction, and the graphs intersect at the moment $t = 4.0\,\text{s}$.

   What does the intersection point tell?

   Choose an option.

   • The accelerations of the cars are then equal, because the lines are at the same height.
   • The velocities of the cars are then equal, but the positions may differ.
   • The cars are then at the same place, because the graphs meet each other.
   • The cars have then travelled the same distance, because they have driven for the same time.

   Answer
   I have done my solution.
   Answer

   The velocities of the cars are then equal, but the positions may differ.

   Show the solution

   Find

   what the intersection point of the $(t,v)$ graphs tells.

   Principle

   A point of a $(t,v)$ graph tells the velocity of the object at the moment $t$. When two graphs intersect, both objects have the same velocity at that moment. The position is determined by all the earlier motion, that is, by the area under the velocity graph, not by a single value of the velocity.

   Model answer

   The velocities of the cars are equal at the moment $4.0\,\text{s}$, but the positions may differ. A meeting cannot be read from this graph, because that needs a $(t,x)$ graph or a comparison of the position functions. The claim that the cars are at the same place mixes up velocity and position. The claim of the same distance travelled does not hold, because the distance is an area and not a single value. The claim of equal accelerations does not hold, because acceleration is the slope, and for intersecting lines the slopes are different.

   Reading the graph

   On the horizontal axis of the graph is time and on the vertical axis velocity. The vertical coordinate of the intersection point is the common velocity of both cars, and the horizontal coordinate is the moment $4.0\,\text{s}$.

7. Exercise 3-7

   Cart A starts from the origin at the moment $t = 0$ and moves at a constant velocity of $4.0\,\text{m/s}$. Cart B starts from the same place $3.0\,\text{s}$ later in the same direction at a constant velocity of $7.0\,\text{m/s}$.

   When does cart B catch up with cart A? Give the time counted from the start of A.

   sCheck
   Answer

   $7.0\,\text{s}$

   Show the solution

   Given

   cart A: $x_\text{A} = v_\text{A}\,t$ when $v_\text{A} = 4.0\,\text{m/s}$; cart B starts with a delay $t_0 = 3.0\,\text{s}$ at a velocity $v_\text{B} = 7.0\,\text{m/s}$; both from the same place

   Find

   the moment $t$ (counted from the start of A) in $\text{s}$

   Principle

   At a meeting the positions are equal. When the clock shows the moment $t$, B has moved only for the time $t - t_0$, so the position of B is $v_\text{B}\,(t - t_0)$.

   Formula

   $v_\text{A}\,t = v_\text{B}\,(t - t_0)$, so

   $(v_\text{B} - v_\text{A})\,t$$= v_\text{B}\,t_0$

   $t$$= \dfrac{v_\text{B}\,t_0}{v_\text{B} - v_\text{A}}$

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   $t$$= \dfrac{7.0\,\text{m/s} \cdot 3.0\,\text{s}}{7.0\,\text{m/s} - 4.0\,\text{m/s}}$

   $= \dfrac{21\,\text{m}}{3.0\,\text{m/s}}$

   $= 7.0\,\text{s}$

   Answer

   $7.0\,\text{s}$

   Counted from the start of A, two significant figures like the given values.

   Reasonableness

   At the moment $7.0\,\text{s}$ A is at the place $4.0\,\text{m/s} \cdot 7.0\,\text{s} = 28\,\text{m}$ and B at the place $7.0\,\text{m/s} \cdot (7.0\,\text{s} - 3.0\,\text{s}) = 28\,\text{m}$. The positions are the same. When B starts, A is already $12\,\text{m}$ ahead, and B approaches at $3.0\,\text{m/s}$, so $12\,\text{m}$ takes $4.0\,\text{s}$, that is, the meeting is at the moment $3.0\,\text{s} + 4.0\,\text{s}$.

8. Exercise 3-8

   Object A accelerates uniformly from rest and object B moves in the same direction at a constant velocity. What condition determines when A passes B?

   Choose an option.

   • The velocities of the objects are equal, because then they travel side by side.
   • The positions of the objects are equal at the same moment, so they are at the same point.
   • The accelerations of the objects are equal, because then the motion is similar.
   • The distances travelled by the objects are equal, even if the starting places were different.

   Answer
   I have done my solution.
   Answer

   The positions of the objects are equal at the same moment, so they are at the same point.

   Show the solution

   Find

   what condition determines the moment of passing.

   Principle

   Passing is a meeting: at the moment of passing, A and B are at the same place. Therefore the position of each is written as a function of time and the positions are set equal, $x_\text{A}(t) = x_\text{B}(t)$. The solution of the equation gives the moment.

   Model answer

   At the moment of passing the positions of the objects are equal at the same moment, that is, they are at the same point. Equal velocities do not determine the passing, because then the lead is at its greatest or smallest, not zero. Equal accelerations are not related to position, and for B the acceleration is zero. Equal distances travelled are not enough if the starting places are different, because the position is the starting place plus the distance travelled.

Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.

1. Exercise 3-9Core

   A bus drives along a straight road at a constant velocity of $72\,\text{km/h}$. A car is $150\,\text{m}$ behind the bus and drives in the same direction at a constant velocity of $90\,\text{km/h}$.

   After how many seconds does the car catch up with the bus?

   sCheck
   Answer

   $30\,\text{s}$

   Show the solution

   Given

   bus $v_\text{bus} = 72\,\text{km/h}$, car $v_\text{car} = 90\,\text{km/h}$, the car is $d = 150\,\text{m}$ behind; both uniform and in the same direction

   Find

   the time $t$ in $\text{s}$

   Principle

   The car catches up with the bus when the positions are equal. Both move uniformly, so $x = x_0 + v\,t$ for each. We choose the origin at the position of the car at the start.

   Formula

   $v_\text{car}\,t = d + v_\text{bus}\,t$, so

   $t$$= \dfrac{d}{v_\text{car} - v_\text{bus}}$

   Conversions

   $90\,\text{km/h}$$= 90 : 3.6\,\text{m/s}$

   $= 25\,\text{m/s}$

   $72\,\text{km/h}$$= 72 : 3.6\,\text{m/s}$

   $= 20\,\text{m/s}$

   Substitution

   $t$$= \dfrac{150\,\text{m}}{25\,\text{m/s} - 20\,\text{m/s}}$

   $= \dfrac{150\,\text{m}}{5.0\,\text{m/s}}$

   $= 30\,\text{s}$

   Answer

   $30\,\text{s}$

   Two significant figures like the velocities.

   Reasonableness

   The difference in velocity is $18\,\text{km/h} = 5.0\,\text{m/s}$, so the car closes in by $5.0\,\text{m}$ every second, and $150\,\text{m}$ takes $30\,\text{s}$. In this time the car drives $25\,\text{m/s} \cdot 30\,\text{s} = 750\,\text{m}$ and the bus $150\,\text{m} + 20\,\text{m/s} \cdot 30\,\text{s} = 750\,\text{m}$.

2. Exercise 3-10Core

   Two trains leave at the same time from stations that are $12\,\text{km}$ apart. The trains move toward each other on parallel tracks at constant velocities of $90\,\text{km/h}$ and $54\,\text{km/h}$.

   After how many minutes do they meet?

   minCheck
   Answer

   $5.0\,\text{min}$

   Show the solution

   Given

   distance between the stations $D = 12\,\text{km}$, velocities $v_1 = 90\,\text{km/h}$ and $v_2 = 54\,\text{km/h}$ toward each other, starting at the same time

   Find

   the time $t$ in $\text{min}$

   Principle

   The trains meet when they are at the same place at the same moment. The origin is at the first station and the positive direction is toward the second station: $x_1 = v_1\,t$ and $x_2 = D - v_2\,t$. The equation $x_1 = x_2$ gives the time.

   Formula

   $v_1\,t = D - v_2\,t$, so $t = \dfrac{D}{v_1 + v_2}$

   Conversions

   We keep the length in kilometres and the velocity in $\text{km/h}$, so the time comes out in hours:

   $t$$= \dfrac{12\,\text{km}}{144\,\text{km/h}}$

   Finally $1\,\text{h} = 60\,\text{min}$.

   Substitution

   $t$$= \dfrac{12\,\text{km}}{90\,\text{km/h} + 54\,\text{km/h}}$

   $= \dfrac{12\,\text{km}}{144\,\text{km/h}}$

   $= 0.0833\,\text{h}$

   $= 0.0833 \cdot 60\,\text{min}$

   $= 5.0\,\text{min}$

   Answer

   $5.0\,\text{min}$

   Two significant figures like the velocities ($54\,\text{km/h}$).

   Reasonableness

   The trains approach each other at $144\,\text{km/h} = 2.4\,\text{km/min}$, so $12\,\text{km}$ takes $12 : 2.4\,\text{min} = 5.0\,\text{min}$. The faster train travels $90\,\text{km/h} \cdot \tfrac{1}{12}\,\text{h} = 7.5\,\text{km}$ and the slower one $4.5\,\text{km}$, together $12\,\text{km}$.

3. Exercise 3-11

   Cart A starts from the place $6.0\,\text{m}$ and moves at a constant velocity of $3.0\,\text{m/s}$ in the positive direction. Cart B starts at the same moment from the place $42\,\text{m}$ and moves at a constant velocity of $-1.0\,\text{m/s}$.

   Sketch on paper the $(t,x)$ graph of each in the same coordinate system and use the graphs to justify why the lines intersect and what the intersection tells. Then state when the carts meet.

   sCheck
   Answer

   $9.0\,\text{s}$

   Show the solution

   Given

   cart A: $x_{0\text{A}} = 6.0\,\text{m}$, $v_\text{A} = 3.0\,\text{m/s}$; cart B: $x_{0\text{B}} = 42\,\text{m}$, $v_\text{B} = -1.0\,\text{m/s}$; both move uniformly

   Find

   the time of the meeting $t$ in $\text{s}$, and a sketch of the graphs with a justification

   Principle

   In uniform motion $x = x_0 + v\,t$, so the $(t,x)$ graph is a straight line: the starting place is where the line begins and the velocity is its slope. The carts meet when they are at the same place at the same moment, that is, when the lines intersect.

   Reading the graph

   The sketch is made like this. Draw time on the horizontal axis and position on the vertical axis. The line of A starts at $6.0\,\text{m}$ and rises $3.0\,\text{m}$ every second. The line of B starts higher, at $42\,\text{m}$, and falls $1.0\,\text{m}$ every second, because the velocity is negative. The rising line starts lower and the falling line higher, so they approach each other and intersect. The intersection gives the time and the place where the carts are at the same point.

   Formula

   $x_\text{A} = x_\text{B}$, that is, $x_{0\text{A}} + v_\text{A}\,t = x_{0\text{B}} + v_\text{B}\,t$. Solving for

   $t$$= \dfrac{x_{0\text{B}} - x_{0\text{A}}}{v_\text{A} - v_\text{B}}$

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   $t$$= \dfrac{42\,\text{m} - 6.0\,\text{m}}{3.0\,\text{m/s} - (-1.0\,\text{m/s})}$

   $= \dfrac{36\,\text{m}}{4.0\,\text{m/s}}$

   $= 9.0\,\text{s}$

   Answer

   $9.0\,\text{s}$

   Two significant figures like the given values.

   Reasonableness

   The position of A at $9.0\,\text{s}$ is $6.0\,\text{m} + 3.0\,\text{m/s} \cdot 9.0\,\text{s} = 33\,\text{m}$ and that of B is $42\,\text{m} - 1.0\,\text{m/s} \cdot 9.0\,\text{s} = 33\,\text{m}$. The positions are equal, so the intersection is $(9.0\,\text{s}$; $33\,\text{m})$, as in the sketch.

4. Exercise 3-12

   The graph shows the positions of two carts as functions of time. The carts move uniformly. From the readings of the graph, calculate the velocities of the carts and then the first moment when the distance between the carts is $10\,\text{m}$.

   sCheck
   Answer

   $3.8\,\text{s}$

   Show the solution

   Given

   from the graph: cart A has $x_0 = 0$ and position $60\,\text{m}$ at $10\,\text{s}$; cart B has $x_0 = 40\,\text{m}$ and position $20\,\text{m}$ at $10\,\text{s}$; the motion is uniform

   Find

   the first moment $t$ when the distance is $10\,\text{m}$, in $\text{s}$

   Principle

   The carts move uniformly, so the graph of each is a straight line and the velocity is the change in position divided by the change in time. Before the meeting B is ahead of A, and the distance is $x_\text{B} - x_\text{A}$. The distance is $10\,\text{m}$ for the first time before the meeting and for the second time after the meeting.

   Reading the graph

   The line of cart A goes from the point $(0\,\text{s}$; $0)$ to the point $(10\,\text{s}$; $60\,\text{m})$, so

   $v_\text{A}$$= \dfrac{60\,\text{m} - 0}{10\,\text{s} - 0}$

   $= 6.0\,\text{m/s}$

   The line of cart B goes from the point $(0\,\text{s}$; $40\,\text{m})$ to the point $(10\,\text{s}$; $20\,\text{m})$, so

   $v_\text{B}$$= \dfrac{20\,\text{m} - 40\,\text{m}}{10\,\text{s} - 0}$

   $= -2.0\,\text{m/s}$

   Formula

   $x_\text{B} - x_\text{A}$$= (40 - 2.0\,t) - 6.0\,t$

   $= 40 - 8.0\,t$

   We set it equal to $10\,\text{m}$:

   $t$$= \dfrac{40\,\text{m} - 10\,\text{m}}{8.0\,\text{m/s}}$

   Conversions

   The positions are in metres and the times in seconds, so no conversions are needed.

   Substitution

   $t$$= \dfrac{40\,\text{m} - 10\,\text{m}}{6.0\,\text{m/s} + 2.0\,\text{m/s}}$

   $= \dfrac{30\,\text{m}}{8.0\,\text{m/s}}$

   $= 3.75\,\text{s}$

   Answer

   $3.8\,\text{s}$

   (two significant figures like the distance $10\,\text{m}$). The unrounded value is $3.75\,\text{s}$.

   Reasonableness

   The distance decreases by $8.0\,\text{m}$ every second and is $40\,\text{m}$ at the start. In $3.75\,\text{s}$ it decreases by $30\,\text{m}$, so $10\,\text{m}$ is left. The moment is before the meeting at $t = 5.0\,\text{s}$, as it should be for the first time.

5. Exercise 3-13

   The positions of two runners on the same straight track were measured by video analysis. Runner A runs in the positive direction and runner B runs toward A. The data are sample data, and both run at a constant velocity.

   Determine the velocities of the runners from the first and the last row of the table and use them to calculate when the runners meet.

   Positions of the runners (sample data)

   $t\,(\text{s})$	$x_\text{A}\,(\text{m})$	$x_\text{B}\,(\text{m})$
   0.0	3.0	45.0
   1.0	6.6	42.4
   2.0	10.1	40.2
   3.0	13.4	37.4
   4.0	17.1	35.1
   5.0	20.4	32.4

   sCheck
   Answer

   $7.0\,\text{s}$

   Show the solution

   Given

   from the table: A: $x_\text{A} = 3.0\,\text{m}$ at $0\,\text{s}$ and $20.4\,\text{m}$ at $5.0\,\text{s}$; B: $x_\text{B} = 45.0\,\text{m}$ at $0\,\text{s}$ and $32.4\,\text{m}$ at $5.0\,\text{s}$; uniform motion

   Find

   the moment of the meeting $t$ in $\text{s}$

   Principle

   In uniform motion the velocity is the change of position divided by the change of time. When the velocities are known, the positions are $x = x_0 + v\,t$, and the runners meet when $x_\text{A} = x_\text{B}$. The points of the table deviate slightly from a straight line because of measurement error, so the first and the last row are used.

   Reading the graph

   The velocities from the end points of the table:

   $v_\text{A}$$= \dfrac{20.4\,\text{m} - 3.0\,\text{m}}{5.0\,\text{s} - 0}$

   $= 3.48\,\text{m/s}$

   $v_\text{B}$$= \dfrac{32.4\,\text{m} - 45.0\,\text{m}}{5.0\,\text{s} - 0}$

   $= -2.52\,\text{m/s}$

   The velocity of B is negative, because B runs toward A.

   Formula

   $x_{0\text{A}} + v_\text{A}\,t = x_{0\text{B}} + v_\text{B}\,t$, so

   $t$$= \dfrac{x_{0\text{B}} - x_{0\text{A}}}{v_\text{A} - v_\text{B}}$

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   $t$$= \dfrac{45.0\,\text{m} - 3.0\,\text{m}}{3.48\,\text{m/s} - (-2.52\,\text{m/s})}$

   $= \dfrac{42.0\,\text{m}}{6.00\,\text{m/s}}$

   $= 7.0\,\text{s}$

   Answer

   $7.0\,\text{s}$

   Two significant figures, because the velocities are determined from three-digit positions but the measurement contains scatter.

   Reasonableness

   The place of the meeting is $3.0\,\text{m} + 3.48\,\text{m/s} \cdot 7.0\,\text{s} = 27.4\,\text{m}$ and the position of B at the same moment is $45.0\,\text{m} - 2.52\,\text{m/s} \cdot 7.0\,\text{s} = 27.4\,\text{m}$. The place is between the starting places, as it should be. The moment $7.0\,\text{s}$ is later than the time interval of the table, so the result is based on extending the line, which is reasonable in uniform motion.

6. Exercise 3-14

   A student solved the following problem. Problem: a cyclist rides at a constant velocity of $4.0\,\text{m/s}$ and at the moment $t = 0$ is at the place $12\,\text{m}$. At the same moment a car at rest at the place $0$ starts accelerating uniformly with an acceleration of $2.0\,\text{m/s}^2$ in the same direction.

   When does the car pass the cyclist?

   The student's solution: “At the moment of passing the velocities are equal, so $a\,t = v$ and $t = 4.0 : 2.0 = 2.0\,\text{s}$.”

   Explain what is wrong in the student's solution and calculate the correct moment of passing. The answer box takes the correct moment of passing.

   sCheck
   Answer

   $6.0\,\text{s}$

   Show the solution

   Given

   cyclist $x_{0\text{B}} = 12\,\text{m}$, $v = 4.0\,\text{m/s}$ (uniform); car $x_0 = 0$, $v_0 = 0$, $a = 2.0\,\text{m/s}^2$

   Find

   what is wrong in the student's solution; the correct moment of passing $t$ in $\text{s}$

   Principle

   Passing is a meeting in which the positions are equal, $x_\text{A} = x_\text{B}$. The position of the car is $\tfrac12\,a\,t^2$ and that of the cyclist $x_{0\text{B}} + v\,t$. The motion of the car is uniformly accelerated, so the equation is a quadratic equation.

   Formula

   $\tfrac12\,a\,t^2 = x_{0\text{B}} + v\,t$, that is, $\tfrac12\,a\,t^2 - v\,t - x_{0\text{B}} = 0$, and the positive root is

   $t$$= \dfrac{v + \sqrt{v^2 + 2\,a\,x_{0\text{B}}}}{a}$

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   The expression under the root is

   $v^2 + 2\,a\,x_{0\text{B}}$$= 16\,\text{m}^2/\text{s}^2 + 48\,\text{m}^2/\text{s}^2$

   $= 64\,\text{m}^2/\text{s}^2$

   So

   $t$$= \dfrac{4.0\,\text{m/s} + 8.0\,\text{m/s}}{2.0\,\text{m/s}^2}$

   $= 6.0\,\text{s}$

   Answer

   $6.0\,\text{s}$

   Two significant figures like the given values.

   Reasonableness

   At the moment $6.0\,\text{s}$ the car is at the place $\tfrac12 \cdot 2.0\,\text{m/s}^2 \cdot (6.0\,\text{s})^2 = 36\,\text{m}$ and the cyclist at the place $12\,\text{m} + 4.0\,\text{m/s} \cdot 6.0\,\text{s} = 36\,\text{m}$. The positions are the same. The velocity of the car is then $12\,\text{m/s}$, three times the velocity of the cyclist, so the car passes as the faster one.

   Model answer on the error

   The student set the velocities equal, but at a meeting it is the positions that are equal, not the velocities. The moment $2.0\,\text{s}$ that the student calculated is the moment when the velocities of the car and the cyclist are equal. Then the car is at the place $4.0\,\text{m}$ and the cyclist at the place $20\,\text{m}$, so the cyclist is $16\,\text{m}$ ahead. This is the greatest value of the lead, not the passing.

7. Exercise 3-15

   Two carts start from the same place in the same direction at the moment $t = 0$. The table gives their velocities (sample data in which the measurement error has been left out). Cart A moves uniformly and cart B accelerates uniformly.

   Draw the $(t,v)$ graphs in the same coordinate system and deduce from the graphs when the distance between the carts is at its greatest. How large is the distance then?

   Velocities of the carts (sample data)

   $t\,(\text{s})$	$v_\text{A}\,(\text{m/s})$	$v_\text{B}\,(\text{m/s})$
   0.0	12.0	0.0
   2.0	12.0	4.0
   4.0	12.0	8.0
   6.0	12.0	12.0
   8.0	12.0	16.0
   10.0	12.0	20.0

   mCheck
   Answer

   $36\,\text{m}$

   Show the solution

   Given

   from the table: A: $v_\text{A} = 12.0\,\text{m/s}$ all the time; B: the velocity increases by $4.0\,\text{m/s}$ in two seconds, $v_\text{B} = 0$ at $0\,\text{s}$ and $20.0\,\text{m/s}$ at $10\,\text{s}$; start from the same place

   Find

   the moment when the distance is at its greatest, and the distance $s_\text{max}$ in $\text{m}$

   Principle

   A is ahead all the time, when B starts from the same place as the slower one. As long as $v_\text{A} > v_\text{B}$, the distance grows, and when B is faster, the distance decreases. So the distance is at its greatest when the velocities are equal. The distance is the area between the graphs up to that moment.

   Reading the graph

   The graph of A is a horizontal line at $12.0\,\text{m/s}$. The graph of B is a straight line rising from the origin with the slope

   $a_\text{B}$$= \dfrac{4.0\,\text{m/s}}{2.0\,\text{s}}$

   $= 2.0\,\text{m/s}^2$

   The lines intersect at the moment $t = 6.0\,\text{s}$, when in the table the velocity of both is $12.0\,\text{m/s}$.

   Formula

   The area is a triangle with the base $t$ and the height $v_\text{A}$: $s_\text{max} = \tfrac12\,t\,v_\text{A}$. The same is obtained from the difference of the positions $v_\text{A}\,t - \tfrac12\,a_\text{B}\,t^2$.

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   $s_\text{max}$$= \tfrac12 \cdot 6.0\,\text{s} \cdot 12.0\,\text{m/s}$

   $= 36\,\text{m}$

   Answer

   The greatest distance is $36\,\text{m}$, and it is at the moment $6.0\,\text{s}$. Two significant figures like the velocities.

   Reasonableness

   From the difference of the positions: A is at the place $12.0\,\text{m/s} \cdot 6.0\,\text{s} = 72\,\text{m}$ and B at the place $\tfrac12 \cdot 2.0\,\text{m/s}^2 \cdot (6.0\,\text{s})^2 = 36\,\text{m}$, so the difference is $36\,\text{m}$. At the moments $4.0\,\text{s}$ and $8.0\,\text{s}$ the difference is $32\,\text{m}$ in both, so $36\,\text{m}$ is the greatest.

Optional: application, two ideas together or a new situation where you reason for yourself.

1. Exercise 3-16

   The bus stop is the origin, and a bus stands there. A runner is behind the stop at the place $-8.0\,\text{m}$ and runs in the positive direction at a constant velocity of $5.0\,\text{m/s}$. At $t = 0$ the bus starts moving and accelerates uniformly at $1.0\,\text{m/s}^2$ in the same direction.

   When does the runner first catch up with the bus?

   sCheck
   Answer

   $2.0\,\text{s}$

   Show the solution

   Given

   runner $x_{0\text{R}} = -8.0\,\text{m}$, $v = 5.0\,\text{m/s}$ (uniform); bus $x_0 = 0$, $v_0 = 0$, $a = 1.0\,\text{m/s}^2$

   Find

   the first moment $t$ when the runner is level with the bus, in $\text{s}$

   Principle

   Meeting: $x_\text{B} = x_\text{R}$. The position of the bus is $\tfrac12\,a\,t^2$ and that of the runner is $x_{0\text{R}} + v\,t$. The equation is quadratic, and both positive roots are meetings: the first time the runner catches up with the bus, and the second time the bus, already faster, passes the runner.

   Formula

   $\tfrac12\,a\,t^2 = x_{0\text{R}} + v\,t$, that is, $\tfrac12\,a\,t^2 - v\,t - x_{0\text{R}} = 0$. The coefficients are $\tfrac12\,a$, $-v$ and $-x_{0\text{R}}$, so the quadratic formula gives

   $t$$= \dfrac{v \pm \sqrt{v^2 + 2\,a\,x_{0\text{R}}}}{a}$

   and the first meeting is the smaller root

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   The expression under the root is

   $v^2 + 2\,a\,x_{0\text{R}}$$= (5.0\,\text{m/s})^2 + 2 \cdot 1.0\,\text{m/s}^2 \cdot (-8.0\,\text{m})$

   $= 9.0\,\text{m}^2/\text{s}^2$

   So

   $t$$= \dfrac{5.0\,\text{m/s} - 3.0\,\text{m/s}}{1.0\,\text{m/s}^2}$

   $= 2.0\,\text{s}$

   Answer

   $2.0\,\text{s}$

   Two significant figures like the given values. The larger root $5.0 + 3.0 = 8.0\,\text{s}$ is the moment when the bus passes the runner.

   Reasonableness

   At $2.0\,\text{s}$ the runner is at $-8.0\,\text{m} + 5.0\,\text{m/s} \cdot 2.0\,\text{s} = 2.0\,\text{m}$, and the bus is at $\tfrac12 \cdot 1.0\,\text{m/s}^2 \cdot (2.0\,\text{s})^2 = 2.0\,\text{m}$. The velocity of the bus is then only $2.0\,\text{m/s}$, so the runner is clearly faster, as is right when catching up.

2. Exercise 3-17

   A bus stands at a stop, and a runner is $20\,\text{m}$ behind it. At $t = 0$ the bus starts moving in the same direction and accelerates uniformly at $0.50\,\text{m/s}^2$. The runner runs in the same direction at a constant velocity of $4.0\,\text{m/s}$.

   Does the runner catch up with the bus? How close to the bus does she get?

   mCheck
   Answer

   $4.0\,\text{m}$

   Show the solution

   Given

   bus $x_{0\text{B}} = 0$, $v_0 = 0$, $a = 0.50\,\text{m/s}^2$; runner $x_{0\text{R}} = -20\,\text{m}$, $v = 4.0\,\text{m/s}$ (uniform); both in the same direction

   Find

   whether the runner catches up with the bus, and the smallest distance $s_\text{min}$ in $\text{m}$

   Principle

   Two ideas: a meeting requires a non-negative expression under the root, and if there is no meeting, the distance is smallest when the velocities are equal (before that the runner is closing in, after that the bus pulls away).

   Formula

   The meeting $\tfrac12\,a\,t^2 = x_{0\text{R}} + v\,t$ gives $\tfrac12\,a\,t^2 - v\,t - x_{0\text{R}} = 0$, and the expression under the root is $v^2 + 2\,a\,x_{0\text{R}}$. The velocities are equal when $a\,t = v$, that is, $t = \dfrac{v}{a}$. The distance is

   $s$$= x_\text{B} - x_\text{R}$

   $= \tfrac12\,a\,t^2 - x_{0\text{R}} - v\,t$

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   The expression under the root:

   $(4.0\,\text{m/s})^2 + 2 \cdot 0.50\,\text{m/s}^2 \cdot (-20\,\text{m})$

   $= 16 - 20$

   $= -4\,\text{m}^2/\text{s}^2$

   $< 0$

   so the runner does not catch up with the bus. The velocities are equal at

   $t$$= \dfrac{4.0\,\text{m/s}}{0.50\,\text{m/s}^2}$

   $= 8.0\,\text{s}$

   Then

   $s_\text{min}$$= \tfrac12 \cdot 0.50\,\text{m/s}^2 \cdot (8.0\,\text{s})^2 + 20\,\text{m} - 4.0\,\text{m/s} \cdot 8.0\,\text{s}$

   $= 16\,\text{m} + 20\,\text{m} - 32\,\text{m}$

   $= 4.0\,\text{m}$

   Answer

   The runner does not catch up with the bus; at closest she is $4.0\,\text{m}$ away, two significant figures like the given values.

   Reasonableness

   At $8.0\,\text{s}$ the bus is at $16\,\text{m}$ and the runner at $-20\,\text{m} + 32\,\text{m} = 12\,\text{m}$, so the gap is $4.0\,\text{m}$. At $6.0\,\text{s}$ and $10\,\text{s}$ the gap is $5.0\,\text{m}$, so $4.0\,\text{m}$ is the smallest. The gap never goes to zero.

3. Exercise 3-18

   A scooter starts from rest with an acceleration of $2.0\,\text{m/s}^2$ toward a cyclist who is $180\,\text{m}$ away and rides toward the scooter at a constant velocity of $8.0\,\text{m/s}$. The scooter and the cyclist use different lanes and meet.

   When are they at the same point?

   sCheck
   Answer

   $10\,\text{s}$

   Show the solution

   Given

   scooter $x_0 = 0$, $v_0 = 0$, $a = 2.0\,\text{m/s}^2$ in the positive direction; cyclist $x_0 = 180\,\text{m}$, velocity $-8.0\,\text{m/s}$ (toward the scooter, uniform)

   Find

   the time of the meeting $t$ in $\text{s}$

   Principle

   The motion of the scooter is uniformly accelerated and that of the cyclist is uniform, and they move in opposite directions. Meeting: $x_\text{S} = x_\text{C}$, when the origin is the starting place of the scooter and the positive direction is toward the cyclist.

   Formula

   $\tfrac12\,a\,t^2 = D - v\,t$, where $D = 180\,\text{m}$ and $v = 8.0\,\text{m/s}$ is the speed of the cyclist. So $\tfrac12\,a\,t^2 + v\,t - D = 0$, and the positive root is

   $t$$= \dfrac{-v + \sqrt{v^2 + 2\,a\,D}}{a}$

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   The expression under the root is

   $v^2 + 2\,a\,D$$= (8.0\,\text{m/s})^2 + 2 \cdot 2.0\,\text{m/s}^2 \cdot 180\,\text{m}$

   $= 784\,\text{m}^2/\text{s}^2$

   So

   $t$$= \dfrac{-8.0\,\text{m/s} + 28\,\text{m/s}}{2.0\,\text{m/s}^2}$

   $= 10\,\text{s}$

   Answer

   $10\,\text{s}$

   Two significant figures like the given values. The negative root $-18\,\text{s}$ would be before the start and does not belong to the situation.

   Reasonableness

   The scooter has driven $\tfrac12 \cdot 2.0\,\text{m/s}^2 \cdot (10\,\text{s})^2 = 100\,\text{m}$ and the cyclist $8.0\,\text{m/s} \cdot 10\,\text{s} = 80\,\text{m}$. Together $100\,\text{m} + 80\,\text{m} = 180\,\text{m}$, so the distance between them has been used up.

4. Exercise 3-19

   Sample situation: a car and a truck drive in the same direction at a velocity of $20\,\text{m/s}$. The length of the truck is $12\,\text{m}$ and that of the car $5.0\,\text{m}$. The front of the car is $10\,\text{m}$ behind the rear of the truck.

   The driver passes the truck. The car accelerates uniformly with an acceleration of $1.5\,\text{m/s}^2$ until the rear of the car is level with the front of the truck, and then it returns to its own lane. The truck continues at a constant velocity. A car comes from the opposite lane at a velocity of $20\,\text{m/s}$, and when the passing begins its front is at a distance $d$ from the front of the car.

   a) How long does the passing take and how far does the car drive during it?

   b) How large must the distance $d$ be at least so that the cars have not had time to meet by the end of the passing? Is $d = 300\,\text{m}$ sufficient, and does the final velocity of the car fit the speed limit $80\,\text{km/h}$?

   The answer box takes the smallest distance $d$ of part b).

   mCheck
   Answer

   $d = 267\,\text{m}$

   Show the solution

   Given

   both $v_0 = 20\,\text{m/s}$; length of the truck $L_\text{T} = 12\,\text{m}$, length of the car $L_\text{c} = 5.0\,\text{m}$, initial gap (front – rear) $10\,\text{m}$; acceleration of the car $a = 1.5\,\text{m/s}^2$; velocity of the oncoming car $v_\text{o} = 20\,\text{m/s}$; speed limit $80\,\text{km/h}$

   Find

   the duration of the passing $t$ and the distance $s_\text{c}$; the smallest distance $d$ in $\text{m}$; the final velocity compared with the limit

   Principle

   Three ideas. First: the motions of the car and the truck are described relative to each other. The truck moves uniformly at $20\,\text{m/s}$ and the car starts from the same velocity, so the difference arises only from the acceleration: the difference of the advance is $\tfrac12\,a\,t^2$. Second: the passing is complete when the difference of the advance is equal to the gap between the front and the rear, the length of the truck and the length of the car together. Third: the cars do not have time to meet if the distances travelled by the car and the oncoming car together are at most $d$.

   Formula

   The relative distance needed is $s_\text{rel} = 10\,\text{m} + L_\text{T} + L_\text{c}$ and $\tfrac12\,a\,t^2 = s_\text{rel}$, so $t = \sqrt{\dfrac{2\,s_\text{rel}}{a}}$. The distance of the car is $s_\text{c} = v_0\,t + \tfrac12\,a\,t^2$, the distance of the oncoming car is $s_\text{o} = v_\text{o}\,t$ and $d_\text{min} = s_\text{c} + s_\text{o}$. The final velocity is $v = v_0 + a\,t$.

   Conversions

   $80\,\text{km/h}$$= 80 : 3.6\,\text{m/s}$

   $= 22.2\,\text{m/s}$

   The other quantities are in SI units.

   Substitution

   $s_\text{rel}$$= 10\,\text{m} + 12\,\text{m} + 5.0\,\text{m}$

   $= 27\,\text{m}$

   So

   $t$$= \sqrt{\dfrac{2 \cdot 27\,\text{m}}{1.5\,\text{m/s}^2}}$

   $= \sqrt{36\,\text{s}^2}$

   $= 6.0\,\text{s}$

   Distance of the car:

   $s_\text{c}$$= 20\,\text{m/s} \cdot 6.0\,\text{s} + \tfrac12 \cdot 1.5\,\text{m/s}^2 \cdot (6.0\,\text{s})^2$

   $= 120\,\text{m} + 27\,\text{m}$

   $= 147\,\text{m}$

   Distance of the oncoming car:

   $s_\text{o}$$= 20\,\text{m/s} \cdot 6.0\,\text{s}$

   $= 120\,\text{m}$

   So

   $d_\text{min}$$= 147\,\text{m} + 120\,\text{m}$

   $= 267\,\text{m}$

   Final velocity:

   $v$$= 20\,\text{m/s} + 1.5\,\text{m/s}^2 \cdot 6.0\,\text{s}$

   $= 29\,\text{m/s}$

   Answer

   The smallest distance is $d = 267\,\text{m}$ (three significant figures, because the result is the sum of two three-digit values). The passing takes $6.0\,\text{s}$, and the car drives $147\,\text{m}$ during it. The distance $300\,\text{m}$ is sufficient by calculation, and a margin of $33\,\text{m}$ remains, but at the end of the passing it is only about $0.7\,\text{s}$ at the common approach velocity of the two cars $\left(33\,\text{m} : 49\,\text{m/s}\right)$. The final velocity $29\,\text{m/s} = 104\,\text{km/h}$ exceeds the limit $80\,\text{km/h}$, so the passing is not allowed even if the distance would be sufficient.

   Model answer on the justification

   The passing should not be done. The calculated margin is too small, because even a small error in the acceleration or in the velocity of the oncoming car uses it up. In addition, the car would have to exceed the speed limit $80\,\text{km/h}$ by about $24\,\text{km/h}$. The model assumes a constant acceleration and exact lengths, so it is rough.

   Reasonableness

   The car has driven $147\,\text{m}$ and the truck $20\,\text{m/s} \cdot 6.0\,\text{s} = 120\,\text{m}$, so the difference is $27\,\text{m}$, as needed. The average velocity of the car is $147\,\text{m} : 6.0\,\text{s} = 24.5\,\text{m/s}$, which is the mean of the initial and final velocities $20\,\text{m/s}$ and $29\,\text{m/s}$.

5. Exercise 3-20

   Sample situation: a cyclist rides at a constant velocity and passes the stop line of a traffic light at the moment $t = 0$. The position $x$ of the cyclist, measured from the stop line, was measured by video analysis (table, sample data).

   A car standing at the stop line starts moving $1.5\,\text{s}$ after the cyclist has passed and accelerates uniformly with an acceleration of $2.0\,\text{m/s}^2$ in the same direction. The area has a speed limit of $40\,\text{km/h}$.

   a) Determine the velocity of the cyclist from the table.

   b) At what place does the car pass the cyclist, and when, counted from the moment the cyclist passed?

   c) What is the velocity of the car at the moment of passing, and does the model obey the speed limit? How would the limit change the result?

   The answer box takes the place of passing in part b).

   Position of the cyclist (sample data)

   $t\,(\text{s})$	$x\,(\text{m})$
   0.0	0.0
   1.0	6.1
   2.0	11.9
   3.0	18.1
   4.0	23.9

   mCheck
   Answer

   $52\,\text{m}$

   Show the solution

   Given

   cyclist: from the table $x = 0$ at $0\,\text{s}$ and $x = 23.9\,\text{m}$ at $4.0\,\text{s}$, uniform motion; the car starts from rest with a delay $t_0 = 1.5\,\text{s}$ at the stop line, $a = 2.0\,\text{m/s}^2$; limit $40\,\text{km/h}$

   Find

   the velocity of the cyclist $v$; the place of passing $x$ (unit $\text{m}$) and the moment of passing $t$; the velocity of the car at the moment of passing compared with the limit

   Principle

   The velocity of the cyclist is obtained from the change of position in the table. The motion of the car begins at the moment $t_0$, so its time of motion is $t - t_0$. At the passing the positions are equal, and the equation is a quadratic equation. Finally the velocity of the car is compared with the limit to see whether the model is reasonable in reality.

   Reading the graph

   The points of the table are almost on a straight line: the position grows by about $6.0\,\text{m}$ every second, so the motion is uniform. From the end points

   $v$$= \dfrac{23.9\,\text{m} - 0}{4.0\,\text{s} - 0}$

   $= 5.975\,\text{m/s}$

   $\approx 6.0\,\text{m/s}$

   Formula

   The cyclist $x_\text{C} = v\,t$ and the car $x_\text{car} = \tfrac12\,a\,(t - t_0)^2$. Let $\tau = t - t_0$, so that $x_\text{C} = v\,(\tau + t_0)$ and at the passing $\tfrac12\,a\,\tau^2 - v\,\tau - v\,t_0 = 0$. The positive root is

   $\tau$$= \dfrac{v + \sqrt{v^2 + 2\,a\,v\,t_0}}{a}$

   Conversions

   $40\,\text{km/h}$$= 40 : 3.6\,\text{m/s}$

   $= 11.1\,\text{m/s}$

   The other quantities are in SI units.

   Substitution

   The expression under the root is

   $v^2 + 2\,a\,v\,t_0$$= 36\,\text{m}^2/\text{s}^2 + 2 \cdot 2.0 \cdot 6.0 \cdot 1.5\,\text{m}^2/\text{s}^2$

   $= 72\,\text{m}^2/\text{s}^2$

   and the square root is $8.49\,\text{m/s}$. So

   $\tau$$= \dfrac{6.0\,\text{m/s} + 8.49\,\text{m/s}}{2.0\,\text{m/s}^2}$

   $= 7.24\,\text{s}$

   $t$$= \tau + t_0$

   $= 7.24\,\text{s} + 1.5\,\text{s}$

   $= 8.74\,\text{s}$

   Place of passing:

   $x$$= v\,t$

   $= 6.0\,\text{m/s} \cdot 8.74\,\text{s}$

   $= 52.4\,\text{m}$

   Velocity of the car at the passing:

   $v_\text{car}$$= a\,\tau$

   $= 2.0\,\text{m/s}^2 \cdot 7.24\,\text{s}$

   $= 14.5\,\text{m/s}$

   Answer

   The car passes the cyclist about $52\,\text{m}$ from the stop line, at the moment $8.7\,\text{s}$ counted from the moment the cyclist passed; two significant figures like the velocity and the acceleration. The velocity of the car is then $14.5\,\text{m/s} \approx 52\,\text{km/h}$, which exceeds the limit $40\,\text{km/h}$. So the model does not obey the limit.

   Model answer on the effect of the limit

   The model assumes that the car accelerates all the time with the same acceleration. The limit $11.1\,\text{m/s}$ is reached already $5.6\,\text{s}$ after the start, before the calculated moment of passing, so the acceleration would end and the car would drive at a constant velocity. Then the passing would happen later and farther than the calculated $52\,\text{m}$, but the car is still faster than the cyclist ($11.1\,\text{m/s} > 6.0\,\text{m/s}$), so it finally passes.

   Reasonableness

   The car has moved $\tfrac12 \cdot 2.0\,\text{m/s}^2 \cdot (7.24\,\text{s})^2 = 52.4\,\text{m}$, the same as the position of the cyclist, so the positions agree. The velocity of the car $14.5\,\text{m/s}$ is greater than that of the cyclist $6.0\,\text{m/s}$, as it should be at a passing.