Position, velocity and uniform motion as graphs

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

1. Exercise 1-1Core

   A runner moves at a uniform velocity along a straight track and covers $120\,\text{m}$ in $15\,\text{s}$ in the direction chosen as positive. What is the runner's velocity?

   m/sCheck
   Answer

   $8.0\,\text{m/s}$

   Show the solution

   Given

   $\Delta x = 120\,\text{m}$, $\Delta t = 15\,\text{s}$; the motion is uniform and in the direction chosen as positive

   Find

   the velocity $v$ of the runner in the unit $\text{m/s}$

   Principle

   The motion is uniform, so the velocity does not change along the whole distance. Then the velocity is the change in position divided by the time it took. The runner moves straight in the positive direction, so the displacement is equal to the distance.

   Formula

   $v = \dfrac{\Delta x}{\Delta t}$. The formula is already solved for the velocity: $v = \dfrac{\Delta x}{\Delta t}$

   Conversions

   The distance is in metres and the time in seconds, so the result comes directly in the unit $\text{m/s}$. No conversion is needed.

   Substitution

   $v$$= \dfrac{120\,\text{m}}{15\,\text{s}}$

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

   Answer

   $8.0\,\text{m/s}$

   Two significant figures like the time $15\,\text{s}$ (the distance $120\,\text{m}$ does not limit the accuracy further).

   Reasonableness

   Estimate: $\dfrac{120\,\text{m}}{15\,\text{s}}$ is $8\,\text{m}$ per second, and in $15\,\text{s}$ $8\,\text{m/s} \cdot 15\,\text{s} = 120\,\text{m}$, so the calculation can be checked backwards. The velocity $8.0\,\text{m/s} \approx 29\,\text{km/h}$ is the speed of a sprinter, so over a distance of $120\,\text{m}$ it is possible. The sign is positive, because the motion is in the positive direction.

2. Exercise 1-2Core

   A cart on the $x$-axis is first at the position $12\,\text{m}$ and then moves to the position $-5.0\,\text{m}$.

   What is the displacement of the cart? Give the answer with its sign.

   mCheck
   Answer

   $-17\,\text{m}$

   Show the solution

   Given

   initial position $x_1 = 12\,\text{m}$, final position $x_2 = -5.0\,\text{m}$

   Find

   the displacement $\Delta x$ with its sign in the unit $\text{m}$

   Principle

   Displacement is the change in position, not the distance travelled. It is always calculated as final position minus initial position, so the sign tells the direction: a negative displacement is motion in the negative direction.

   Formula

   $\Delta x = x_2 - x_1$. The formula is already solved for the displacement: $\Delta x = x_2 - x_1$

   Conversions

   Both positions are in metres, so no units need to be converted.

   Substitution

   $\Delta x$$= -5.0\,\text{m} - 12\,\text{m}$

   $= -17\,\text{m}$

   Answer

   $-17\,\text{m}$

   Two significant figures. The position $12\,\text{m}$ is known to the accuracy of a metre, so the difference has that accuracy.

   Reasonableness

   The cart moved in the negative direction, so the displacement must be negative. The cart first went $12\,\text{m}$ to the origin and then $5.0\,\text{m}$ past it, altogether $12\,\text{m} + 5.0\,\text{m} = 17\,\text{m}$. Here the distance and the magnitude of the displacement are equal, as long as the cart does not turn around.

3. Exercise 1-3

   A cart moves at a uniform velocity of $3.5\,\text{m/s}$ in the positive direction for $12\,\text{s}$. How large is the displacement of the cart?

   mCheck
   Answer

   $42\,\text{m}$

   Show the solution

   Given

   $v = 3.5\,\text{m/s}$, $\Delta t = 12\,\text{s}$; the motion is uniform and in the positive direction

   Find

   the displacement $\Delta x$ in the unit $\text{m}$

   Principle

   In uniform motion the velocity stays constant, so the position changes by the same amount every second. The displacement is therefore the velocity multiplied by the time.

   Formula

   $v = \dfrac{\Delta x}{\Delta t}$, so $\Delta x = v\,\Delta t$

   Conversions

   The velocity is in the unit $\text{m/s}$ and the time in seconds, so the seconds cancel and the result is in metres. No conversion is needed.

   Substitution

   $\Delta x$$= 3.5\,\text{m/s} \cdot 12\,\text{s}$

   $= 42\,\text{m}$

   Answer

   $42\,\text{m}$

   Two significant figures like both initial values ($3.5\,\text{m/s}$ and $12\,\text{s}$).

   Reasonableness

   The unit is correct: $\text{m/s} \cdot \text{s} = \text{m}$. Estimate: about $3\,\text{m/s} \cdot 12\,\text{s} = 36\,\text{m}$, so $42\,\text{m}$ is of the right order of magnitude. The velocity $3.5\,\text{m/s}$ corresponds to about $13\,\text{km/h}$, that is jogging, and in $12\,\text{s}$ a jogger covers a few tens of metres. The displacement is positive, because the motion is in the positive direction.

4. Exercise 1-4

   The $(t, x)$ graph of a cart is a straight line that passes through the points $(0\,\text{s};\, 4.0\,\text{m})$ and $(10\,\text{s};\, 24\,\text{m})$. What is the velocity of the cart?

   m/sCheck
   Answer

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

   Show the solution

   Given

   points of the line $(t_1$; $x_1) = (0\,\text{s}$; $4.0\,\text{m})$ and $(t_2$; $x_2) = (10\,\text{s}$; $24\,\text{m})$

   Find

   the velocity $v$ of the cart in the unit $\text{m/s}$

   Principle

   In uniform motion the velocity is constant, so the $(t, x)$ graph is a straight line. The slope of the line, that is the rise divided by the horizontal run, is the change in position divided by the change in time, that is the velocity.

   Reading the graph

   We use the two given points of the line: the position of the first is $x_1 = 4.0\,\text{m}$ at the moment $t_1 = 0\,\text{s}$ and the position of the second is $x_2 = 24\,\text{m}$ at the moment $t_2 = 10\,\text{s}$. The vertical rise is $x_2 - x_1$ and the horizontal run is $t_2 - t_1$. Their ratio is the velocity.

   Formula

   $v = \dfrac{\Delta x}{\Delta t}$, where $\Delta x = x_2 - x_1$ and $\Delta t = t_2 - t_1$. So $v = \dfrac{x_2 - x_1}{t_2 - t_1}$

   Conversions

   The positions are in metres and the times in seconds, so no conversions are needed. The starting position $4.0\,\text{m}$ is not zero, but it cancels in the difference.

   Substitution

   $v$$= \dfrac{24\,\text{m} - 4.0\,\text{m}}{10\,\text{s} - 0\,\text{s}}$

   $= \dfrac{20\,\text{m}}{10\,\text{s}}$

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

   Answer

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

   Two significant figures, because the position $24\,\text{m}$ is given to the accuracy of a metre ($4.0\,\text{m}$ to a decimetre), so the difference $20\,\text{m}$ has two figures; the time $10\,\text{s}$ also has two.

   Reasonableness

   The graph rises, so the velocity is positive. Check: $4.0\,\text{m} + 2.0\,\text{m/s} \cdot 10\,\text{s} = 24\,\text{m}$, which is the other point of the line. If the velocity were calculated as

   $\dfrac{24\,\text{m}}{10\,\text{s}}$$= 2.4\,\text{m/s}$

   the starting position would be forgotten.

5. Exercise 1-5

   At the moment $t = 0\,\text{s}$ a walker is at the position $15\,\text{m}$ and walks uniformly at a velocity $1.5\,\text{m/s}$ in the positive direction.

   At what position is she at the moment $t = 20\,\text{s}$?

   mCheck
   Answer

   $45\,\text{m}$

   Show the solution

   Given

   $x_0 = 15\,\text{m}$ at the moment $t = 0\,\text{s}$, $v = 1.5\,\text{m/s}$, the moment asked for $t = 20\,\text{s}$

   Find

   the position $x$ at the moment $t = 20\,\text{s}$, in the unit $\text{m}$

   Principle

   The motion is uniform, so the position changes by the product of the velocity and the time. The position is the initial position plus this displacement, because the walker did not start from the origin.

   Formula

   $x = x_0 + v\,t$. The formula is already solved for the position: $x = x_0 + v\,t$

   Conversions

   The position is in metres, the velocity in the unit $\text{m/s}$ and the time in seconds, so no conversions are needed.

   Substitution

   $x$$= 15\,\text{m} + 1.5\,\text{m/s} \cdot 20\,\text{s}$

   $= 15\,\text{m} + 30\,\text{m}$

   $= 45\,\text{m}$

   Answer

   $45\,\text{m}$

   Two significant figures, because the velocity $1.5\,\text{m/s}$ and the time $20\,\text{s}$ have two figures and the sum $15\,\text{m} + 30\,\text{m}$ is to the accuracy of a metre.

   Reasonableness

   The walker moves $30\,\text{m}$ forward, so the position must be greater than $15\,\text{m}$, and $45\,\text{m}$ is. The velocity $1.5\,\text{m/s} \approx 5.4\,\text{km/h}$ is a normal walking speed. The product $v\,t = 30\,\text{m}$ alone would be the displacement, not the position. The limiting case $t = 0$ gives $x = x_0 = 15\,\text{m}$, as it should.

6. Exercise 1-6

   The $(t, x)$ graph of an object is a straight line that falls from left to right. What does this tell about the motion of the object?

   Choose an option.

   • The object is moving downhill, because the graph goes downwards and the graph shows the shape of the path.
   • The object is coming to a stop, because the line of the graph tilts and the magnitude of the velocity decreases.
   • The position decreases as time passes, so the velocity is negative and the motion is in the negative direction.
   • As time passes the object moves backwards in time, because the position values of the graph get smaller.

   Answer
   I have done my solution.
   Answer

   The position decreases as time passes, so the velocity is negative and the motion is in the negative direction.

   Show the solution

   Find

   what a falling line in a $(t, x)$ graph tells about the motion.

   Principle

   The slope of a line in a $(t, x)$ graph is the velocity $v = \dfrac{\Delta x}{\Delta t}$. When the line falls, the position decreases as time passes, so the slope and the velocity are negative. The graph shows position as a function of time, not the shape of the path.

   Model answer

   The position decreases as time passes, so the velocity is negative and the object moves in the negative direction. The option in which the object moves downhill confuses the graph with a picture of the path: the object can move along a horizontal track even though the graph falls. A stopped object is at rest, and then the graph is horizontal. A straight line whose slope does not change describes a constant velocity, not a slowing motion. Time always increases as one goes to the right, so there is no object travelling backwards in time.

   Reading the graph

   In a falling line $x$ is smaller when $t$ is greater, so $\Delta x < 0$ and $\Delta t > 0$. Then $\dfrac{\Delta x}{\Delta t} < 0$.

7. Exercise 1-7

   A walker walks $40\,\text{m}$ east along a straight road, turns around and walks the same $40\,\text{m}$ back to the starting point. East has been chosen as the positive direction.

   Which statement about the whole walk is correct?

   Choose an option.

   • The displacement is $40\,\text{m}$, because only the outward leg counts when the motion turns around.
   • The displacement is $-40\,\text{m}$, because the last leg goes in the negative direction.
   • The displacement is $80\,\text{m}$, because the displacement is the whole length of the route travelled.
   • The displacement is $0\,\text{m}$, because the final position is the same as the initial position.

   Answer
   I have done my solution.
   Answer

   The displacement is $0\,\text{m}$, because the final position is the same as the initial position.

   Show the solution

   Find

   which statement about the displacement of the whole walk is correct.

   Principle

   The displacement is the final position minus the initial position, $\Delta x = x_2 - x_1$, and it does not depend on the route. The distance is the length of the route travelled. Partial displacements in different directions add up with their signs.

   Model answer

   The walker returns to the starting point, so the final position is the same as the initial position and the displacement is $0\,\text{m}$. The same result is obtained by adding the partial displacements: $+40\,\text{m} + (-40\,\text{m}) = 0\,\text{m}$. The distance travelled is $40\,\text{m} + 40\,\text{m} = 80\,\text{m}$, but the distance is not the displacement. The statement that the displacement is $-40\,\text{m}$ would be true only for the last leg, not for the whole walk. The statement that only the outward leg counts forgets the return, which cancels the outward leg.

8. Exercise 1-8

   A student calculates the velocity of a cart from a straight line of the $(t, x)$ graph, whose points are $(2.0\,\text{s};\, 5.0\,\text{m})$ and $(6.0\,\text{s};\, 13\,\text{m})$.

   The student's solution: $v = \dfrac{13\,\text{m}}{6.0\,\text{s}} = 2.2\,\text{m/s}$.

   What is the error in the student's solution?

   Choose an option.

   • There is no error, because the velocity is the final position divided by the total time elapsed since the beginning.
   • She divided the position of the second point by its time, although one must divide the change in position by the change in time.
   • She calculated speed and not velocity, because velocity is obtained from the sum of the positions divided by the sum of the times.
   • She swapped the points, so the sign is wrong and the correct velocity is $-2.2\,\text{m/s}$.

   Answer
   I have done my solution.
   Answer

   She divided the position of the second point by its time, although one must divide the change in position by the change in time.

   Show the solution

   Find

   what the error is in the student's calculation of the velocity.

   Principle

   The slope of a line is the change in position divided by the change in time, $v = \dfrac{x_2 - x_1}{t_2 - t_1}$. A change is always a difference, final minus initial, so the position of the second point alone is not valid as the numerator and the time of the second point alone is not valid as the denominator.

   Model answer

   The student divided the position of the second point, $13\,\text{m}$, by its time, $6.0\,\text{s}$, and did not calculate the differences. Correctly: $v = \dfrac{13\,\text{m} - 5.0\,\text{m}}{6.0\,\text{s} - 2.0\,\text{s}} = \dfrac{8.0\,\text{m}}{4.0\,\text{s}} = 2.0\,\text{m/s}$. The student's method would work only if the line started from the origin at the moment zero, but here the starting point is $(2.0\,\text{s}$; $5.0\,\text{m})$. The sign is not wrong, because the position increases and the velocity is positive. The velocity is not the sum of the positions divided by the sum of the times, because a sum does not describe a change.

   Reading the graph

   Two points are read from the line: $(t_1$; $x_1) = (2.0\,\text{s}$; $5.0\,\text{m})$ and $(t_2$; $x_2) = (6.0\,\text{s}$; $13\,\text{m})$. The rise is $x_2 - x_1$ and the run is $t_2 - t_1$.

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

1. Exercise 1-9Core

   A car drives along a straight road at a uniform velocity $90\,\text{km/h}$. How far does it travel in $12\,\text{s}$? Give the answer in metres.

   mCheck
   Answer

   $300\,\text{m}$

   Show the solution

   Given

   $v = 90\,\text{km/h}$, $\Delta t = 12\,\text{s}$; the motion is uniform and straight

   Find

   the displacement $\Delta x$ in metres; the car drives along a straight road in one direction, so the distance $s = v\,\Delta t$ is equal

   Principle

   The car drives at a uniform velocity along a straight road in one direction, so the displacement is the velocity times the time. The velocity and the time must first be put into the same system of units, so that the result comes in metres.

   Formula

   $v = \dfrac{\Delta x}{\Delta t}$, so $\Delta x = v\,\Delta t$

   Conversions

   One $\text{km/h}$ is

   $\dfrac{1000\,\text{m}}{3600\,\text{s}}$$= \dfrac{1}{3.6}\,\text{m/s}$

   so

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

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

   Substitution

   $\Delta x$$= 25\,\text{m/s} \cdot 12\,\text{s}$

   $= 300\,\text{m}$

   Answer

   $300\,\text{m}$

   Or $3.0 \cdot 10^2\,\text{m}$, two significant figures, because the velocity $90\,\text{km/h}$ and the time $12\,\text{s}$ have two figures; the conversion factor $3.6$ is exact and does not limit the accuracy.

   Reasonableness

   The car travels $25\,\text{m}$ every second, which is about the length of six cars. In ten seconds the distance would be $250\,\text{m}$ and in twelve a little more, so $300\,\text{m}$ fits. If one calculated directly $90 \cdot 12 = 1080$, one would mix up the units: that would be $\text{km/h} \cdot \text{s}$, not metres.

2. Exercise 1-10Core

   A runner moves along a straight path at a uniform velocity of $4.0\,\text{m/s}$. How many minutes does it take her to cover a distance of $1.2\,\text{km}$?

   minCheck
   Answer

   $5.0\,\text{min}$

   Show the solution

   Given

   $v = 4.0\,\text{m/s}$, $\Delta x = 1.2\,\text{km}$; the motion is uniform and straight

   Find

   the time $\Delta t$ in minutes

   Principle

   The velocity is constant, so the distance is covered at the same rate all the time. The time can be solved from the definition of velocity when the distance and the velocity are known.

   Formula

   $v = \dfrac{\Delta x}{\Delta t}$, so $\Delta t = \dfrac{\Delta x}{v}$

   Conversions

   $1.2\,\text{km}$$= 1.2 \cdot 1000\,\text{m}$

   $= 1200\,\text{m}$

   so that the units are $\text{m}$ and $\text{m/s}$. The time comes out in seconds, and $1\,\text{min} = 60\,\text{s}$, so

   $\Delta t_\text{min}$$= \dfrac{\Delta t}{60}$

   Substitution

   $\Delta t$$= \dfrac{1200\,\text{m}}{4.0\,\text{m/s}}$

   $= 300\,\text{s}$

   $= \dfrac{300}{60}\,\text{min}$

   $= 5.0\,\text{min}$

   Answer

   $5.0\,\text{min}$

   Two significant figures like the velocity $4.0\,\text{m/s}$ and the distance $1.2\,\text{km}$.

   Reasonableness

   The velocity $4.0\,\text{m/s}$ is $14.4\,\text{km/h}$, so a kilometre takes about $4.2\,\text{min}$, and $1.2\,\text{km}$ takes about $1.2 \cdot 4.2\,\text{min} \approx 5\,\text{min}$, so $5.0\,\text{min}$ fits. The time is positive, as it should be.

3. Exercise 1-11

   At the moment $t = 2.0\,\text{s}$ a cart is at the position $18\,\text{m}$ and at the moment $t = 7.0\,\text{s}$ at the position $3.0\,\text{m}$. The cart moves uniformly. Calculate the velocity of the cart with its sign.

   m/sCheck
   Answer

   $-3.0\,\text{m/s}$

   Show the solution

   Given

   $(t_1$; $x_1) = (2.0\,\text{s}$; $18\,\text{m})$ and $(t_2$; $x_2) = (7.0\,\text{s}$; $3.0\,\text{m})$; the motion is uniform

   Find

   the velocity $v$ with its sign in the unit $\text{m/s}$

   Principle

   The motion is uniform, so the average velocity between two moments is the same as the velocity at all times. The velocity is calculated from the change in position divided by the change in time, and the sign tells the direction.

   Formula

   $v = \dfrac{\Delta x}{\Delta t}$, where $\Delta x = x_2 - x_1$ and $\Delta t = t_2 - t_1$. So $v = \dfrac{x_2 - x_1}{t_2 - t_1}$

   Conversions

   The positions are in metres and the times in seconds, so no conversions are needed. The differences are always calculated as final minus initial.

   Substitution

   $v$$= \dfrac{3.0\,\text{m} - 18\,\text{m}}{7.0\,\text{s} - 2.0\,\text{s}}$

   $= \dfrac{-15\,\text{m}}{5.0\,\text{s}}$

   $= -3.0\,\text{m/s}$

   Answer

   $-3.0\,\text{m/s}$

   Two significant figures, because the difference of the positions $3.0\,\text{m} - 18\,\text{m} = -15\,\text{m}$ is to the accuracy of a metre (two figures) and the change in time $5.0\,\text{s}$ also has two.

   Reasonableness

   The position decreases, so the velocity must be negative: the cart moves towards the origin. Check backwards: $18\,\text{m} + (-3.0\,\text{m/s}) \cdot 5.0\,\text{s} = 3.0\,\text{m}$, which is the other given position. The speed is $3.0\,\text{m/s}$.

4. Exercise 1-12

   At the moment $t = 0\,\text{s}$ a cart is at the position $-24\,\text{m}$ and moves uniformly at a velocity $1.5\,\text{m/s}$ in the positive direction.

   When is the cart at the origin, that is at the position $x = 0$?

   sCheck
   Answer

   $16\,\text{s}$

   Show the solution

   Given

   $x_0 = -24\,\text{m}$ at the moment $t = 0\,\text{s}$, $v = 1.5\,\text{m/s}$, the position asked for $x = 0$

   Find

   the time $t$ in seconds

   Principle

   The motion is uniform, so the position follows the equation $x = x_0 + v\,t$. The cart is at the origin when its position is zero, and the time can be solved from this condition.

   Formula

   $x = x_0 + v\,t$, so $v\,t = x - x_0$ and $t = \dfrac{x - x_0}{v}$

   Conversions

   The positions are in metres and the velocity in the unit $\text{m/s}$, so the time comes directly in seconds. No conversions are needed.

   Substitution

   $t$$= \dfrac{0 - (-24\,\text{m})}{1.5\,\text{m/s}}$

   $= \dfrac{24\,\text{m}}{1.5\,\text{m/s}}$

   $= 16\,\text{s}$

   Answer

   $16\,\text{s}$

   Two significant figures like the initial values $24\,\text{m}$ and $1.5\,\text{m/s}$.

   Reasonableness

   The cart is on the negative side of the origin and moves in the positive direction, so it arrives at the origin, and the time is positive. Check backwards:

   $-24\,\text{m} + 1.5\,\text{m/s} \cdot 16\,\text{s}$

   $= -24\,\text{m} + 24\,\text{m}$

   $= 0$

   Estimate: $1.5\,\text{m/s}$ is about $1\,\text{m}$ for every 0.7 seconds, and $24\,\text{m}$ takes about $16\,\text{s}$.

5. Exercise 1-13

   The motion of a cart was followed with video analysis. The table gives the position of the cart at different moments (example data). The motion of the cart is uniform.

   Draw the $(t, x)$ graph from the data and fit a straight line through the points. Use the line to determine the velocity $v$ of the cart. Also give the initial position $x_0$ and write the position of the cart as a function of time.

   Write the velocity in the answer box.

   Position of the cart in video analysis (example data)

   $t\,(\text{s})$	$x\,(\text{m})$
   0.0	0.48
   1.0	1.12
   2.0	1.67
   3.0	2.33
   4.0	2.88
   5.0	3.53

   m/sCheck
   Answer

   $0.61\,\text{m/s}$

   Show the solution

   Given

   the points $(t$; $x)$ of the table, starting from $(0.0\,\text{s}$; $0.48\,\text{m})$ and ending at $(5.0\,\text{s}$; $3.53\,\text{m})$; the motion is uniform

   Find

   the velocity $v$ in the unit $\text{m/s}$, the initial position $x_0$ and the position as a function of time

   Principle

   In uniform motion the $(t, x)$ points lie on a line. The measurements contain small errors, so the line is fitted through the points so that the points are evenly on both sides of it. The velocity is the slope of the line, and the initial position is the intercept of the line on the position axis.

   Reading the graph

   The time axis is $t = 0 \ldots 5.0\,\text{s}$ and the position axis $0 \ldots 4\,\text{m}$. The points are marked and the line is drawn through them. The line passes close to the points $(0.0\,\text{s}$; $0.49\,\text{m})$ and $(5.0\,\text{s}$; $3.54\,\text{m})$. The slope is calculated from these points of the line that are far apart, not from adjacent measurement points.

   Formula

   $v = \dfrac{x_2 - x_1}{t_2 - t_1}$ and $x = x_0 + v\,t$

   Conversions

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

   Substitution

   $$\begin{aligned} v &= \dfrac{3.54\,\text{m} - 0.49\,\text{m}}{5.0\,\text{s} - 0.0\,\text{s}} \\ &= 0.61\,\text{m/s} \end{aligned}$$

   The initial position is read from the intercept of the line: $x_0 \approx 0.49\,\text{m}$. So the position is $x = 0.49\,\text{m} + 0.61\,\text{m/s} \cdot t$.

   Answer

   $0.61\,\text{m/s}$

   Two significant figures, because the positions of the table are given to two decimals and the fitting of the line limits the accuracy. The reading of the line can vary by a few per cent.

   Reasonableness

   Check with the middle point: $0.49\,\text{m} + 0.61\,\text{m/s} \cdot 2.0\,\text{s} = 1.71\,\text{m}$, while the table has $1.67\,\text{m}$, so the difference is small. The position increases, so the velocity is positive. The velocity $0.61\,\text{m/s}$ suits a small cart.

6. Exercise 1-14

   Two carts move along the same straight track and both motions are uniform. The table gives the positions of carts A and B at different moments (example data).

   Draw the $(t, x)$ points of both carts in the same graph and fit a straight line to each. Determine the velocities of the carts from the lines and tell which one moves faster.

   When are the carts at the same position? Write this moment in the answer box.

   Positions of carts A and B (example data)

   $t\,(\text{s})$	$x_\text{A}\,(\text{m})$	$x_\text{B}\,(\text{m})$
   0.0	1.0	20.1
   2.0	4.1	16.4
   4.0	6.9	12.9
   6.0	10.1	9.2
   8.0	13.0	5.8

   sCheck
   Answer

   $5.8\,\text{s}$

   Show the solution

   Given

   the positions in the table for carts A and B at the moments $0.0\,\text{s} \ldots 8.0\,\text{s}$; the motions are uniform

   Find

   which cart is faster and the moment $t$ when the carts are at the same position, in the unit $\text{s}$

   Principle

   The points of each cart lie on a line. The slope of the line is the velocity and the intercept is the initial position. The carts are at the same position when the lines intersect, that is, when the positions are equal at the same moment.

   Reading the graph

   A line is fitted for A close to the points $(0.0\,\text{s}$; $1.0\,\text{m})$ and $(8.0\,\text{s}$; $13.0\,\text{m})$ and for B close to the points $(0.0\,\text{s}$; $20.0\,\text{m})$ and $(8.0\,\text{s}$; $5.7\,\text{m})$. From these we get

   $v_\text{A}$$= \dfrac{13.0\,\text{m} - 1.0\,\text{m}}{8.0\,\text{s}}$

   $= 1.5\,\text{m/s}$

   $v_\text{B}$$= \dfrac{5.7\,\text{m} - 20.0\,\text{m}}{8.0\,\text{s}}$

   $\approx -1.8\,\text{m/s}$

   and the initial positions are $x_{0\text{A}} = 1.0\,\text{m}$ and $x_{0\text{B}} = 20.0\,\text{m}$. The speed of cart B is greater ($1.8\,\text{m/s} > 1.5\,\text{m/s}$), but it moves in the negative direction toward A.

   Formula

   $x_\text{A} = x_{0\text{A}} + v_\text{A}\,t$ and $x_\text{B} = x_{0\text{B}} + v_\text{B}\,t$. From the condition $x_\text{A} = x_\text{B}$ we get
   $$t = \dfrac{x_{0\text{B}} - x_{0\text{A}}}{v_\text{A} - v_\text{B}}$$

   Conversions

   All quantities are in the units $\text{m}$ and $\text{m/s}$, so no conversions are needed. The velocity of B is substituted as negative.

   Substitution

   $$\begin{aligned} t &= \dfrac{20.0\,\text{m} - 1.0\,\text{m}}{1.5\,\text{m/s} - (-1.8\,\text{m/s})} \\ &= \dfrac{19.0\,\text{m}}{3.3\,\text{m/s}} \\ &= 5.8\,\text{s} \end{aligned}$$

   Answer

   $5.8\,\text{s}$

   Two significant figures, because the velocities $1.5\,\text{m/s}$ and $1.8\,\text{m/s}$ have two figures.

   Reasonableness

   At the moment $5.8\,\text{s}$ cart A is at the position $1.0\,\text{m} + 1.5\,\text{m/s} \cdot 5.8\,\text{s} = 9.7\,\text{m}$ and cart B at the position $20.0\,\text{m} - 1.8\,\text{m/s} \cdot 5.8\,\text{s} = 9.6\,\text{m}$, so the positions are equal to within the rounding accuracy. The moment is inside the measurement interval $0.0\ldots 8.0\,\text{s}$, and in the table the order of the carts changes between the moments $4.0\,\text{s}$ and $6.0\,\text{s}$, as it should.

7. Exercise 1-15

   A robot moves along a straight track, and its position has been measured at certain moments (example data). Between two measurement points the motion is uniform.

   Draw the $(t, x)$ graph of the robot and, based on it, the $(t, v)$ graph by drawing the velocity of each time interval. During which time interval does the robot move in the negative direction?

   Write in the answer box the velocity of the robot during this interval, with its sign.

   Position of the robot (example data)

   $t\,(\text{s})$	$x\,(\text{m})$
   0.0	2.0
   4.0	14
   10	14
   16	5.0
   20	5.0

   m/sCheck
   Answer

   $-1.5\,\text{m/s}$

   Show the solution

   Given

   the points $(t$; $x)$ of the robot: $(0\,\text{s}$; $2.0\,\text{m})$, $(4.0\,\text{s}$; $14\,\text{m})$, $(10\,\text{s}$; $14\,\text{m})$, $(16\,\text{s}$; $5.0\,\text{m})$ and $(20\,\text{s}$; $5.0\,\text{m})$; between the points the motion is uniform

   Find

   the time interval during which the robot moves in the negative direction, and the velocity $v$ during that interval in the unit $\text{m/s}$

   Principle

   In each time interval the $(t, x)$ graph is a line, and its slope is the velocity. A falling part gives a negative velocity, a horizontal part a velocity of zero and a rising part a positive velocity. So the $(t, v)$ graph consists of horizontal lines.

   Reading the graph

   The $(t, x)$ graph has four parts: a rising interval $0\ldots4.0\,\text{s}$, a horizontal interval $4.0\ldots10\,\text{s}$, a falling interval $10\ldots16\,\text{s}$ and a horizontal interval $16\ldots20\,\text{s}$. The falling interval is $10\ldots16\,\text{s}$. The velocity of the first interval is

   $\dfrac{14\,\text{m} - 2.0\,\text{m}}{4.0\,\text{s}}$$= 3.0\,\text{m/s}$

   In the velocity graph the velocity is $3.0\,\text{m/s}$ during $0\ldots4.0\,\text{s}$, zero during $4.0\ldots10\,\text{s}$, negative during $10\ldots16\,\text{s}$ and zero during $16\ldots20\,\text{s}$.

   Formula

   $v = \dfrac{x_2 - x_1}{t_2 - t_1}$

   Conversions

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

   Substitution

   $$\begin{aligned} v &= \dfrac{5.0\,\text{m} - 14\,\text{m}}{16\,\text{s} - 10\,\text{s}} \\ &= \dfrac{-9.0\,\text{m}}{6.0\,\text{s}} \\ &= -1.5\,\text{m/s} \end{aligned}$$

   Answer

   $-1.5\,\text{m/s}$

   During $10\ldots16\,\text{s}$, two significant figures like the difference of the positions, $9.0\,\text{m}$.

   Reasonableness

   The position of the robot decreases during $10\ldots16\,\text{s}$, so the velocity must be negative. Checking backwards: $14\,\text{m} + (-1.5\,\text{m/s}) \cdot 6.0\,\text{s} = 5.0\,\text{m}$, which is the position of the table at the moment $16\,\text{s}$. In the horizontal parts the velocity is zero, because the position does not change.

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

1. Exercise 1-16

   A cyclist rides east along a straight road $6.0\,\text{km}$ in $20\,\text{min}$ and then turns straight back, riding the same $6.0\,\text{km}$ west in $30\,\text{min}$. Calculate the average speed over the whole trip, that is the distance travelled divided by the elapsed time. Give the answer in the unit m/s.

   m/sCheck
   Answer

   $4.0\,\text{m/s}$

   Show the solution

   Given

   east $s_1 = 6.0\,\text{km}$ in the time $\Delta t_1 = 20\,\text{min}$ and west $s_2 = 6.0\,\text{km}$ in the time $\Delta t_2 = 30\,\text{min}$

   Find

   the average speed $v_\text{speed}$ in the unit $\text{m/s}$

   Principle

   The average speed is the whole distance travelled divided by the elapsed time. Both parts are included in the distance, because the distance does not depend on the direction. The average velocity would be a different thing, because it would use the displacement, in which the direction matters.

   Formula

   $v_\text{speed} = \dfrac{s}{\Delta t}$, where $s = s_1 + s_2$ and $\Delta t = \Delta t_1 + \Delta t_2$. So

   $v_\text{speed}$$= \dfrac{s_1 + s_2}{\Delta t_1 + \Delta t_2}$

   Conversions

   $6.0\,\text{km} = 6000\,\text{m}$ and $20\,\text{min} = 1200\,\text{s}$, $30\,\text{min} = 1800\,\text{s}$.

   Substitution

   $v_\text{speed}$$= \dfrac{6000\,\text{m} + 6000\,\text{m}}{1200\,\text{s} + 1800\,\text{s}}$

   $= \dfrac{12\,000\,\text{m}}{3000\,\text{s}}$

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

   Answer

   $4.0\,\text{m/s}$

   Two significant figures, because the times $20\,\text{min}$ and $30\,\text{min}$ (the sum $50\,\text{min}$) have two figures; the distances $6.0\,\text{km}$ also have two, and their sum $12.0\,\text{km}$ limits less.

   Reasonableness

   The speed on the way out is

   $\dfrac{6000\,\text{m}}{1200\,\text{s}}$$= 5.0\,\text{m/s}$

   and the speed on the way back

   $\dfrac{6000\,\text{m}}{1800\,\text{s}}$$\approx 3.3\,\text{m/s}$

   and the value lies between them, as an average speed should. The cyclist returned to the starting point, so the displacement is $0$ and the average velocity $0\,\text{m/s}$, even though the average speed is not zero. About $14\,\text{km/h}$ is relaxed cycling.

2. Exercise 1-17

   Aino and Eeva start walking towards each other at the same time along the same line. Aino starts from the position 0 and walks at a velocity $1.5\,\text{m/s}$ in the positive direction. Eeva starts from the position $90\,\text{m}$ and walks at a velocity $-1.2\,\text{m/s}$.

   At what position do they meet?

   mCheck
   Answer

   $50\,\text{m}$

   Show the solution

   Given

   Aino: $x_{0\text{A}} = 0$, $v_\text{A} = 1.5\,\text{m/s}$; Eeva: $x_{0\text{E}} = 90\,\text{m}$, $v_\text{E} = -1.2\,\text{m/s}$; both start at the same time and the motion is uniform

   Find

   the meeting position $x$ in the unit $\text{m}$

   Principle

   The motion of both is uniform, so the position of each is of the form $x_0 + v\,t$. They meet when their positions are the same at the same moment, and the time is solved first from this condition.

   Formula

   $x_\text{A} = x_{0\text{A}} + v_\text{A}\,t$ and $x_\text{E} = x_{0\text{E}} + v_\text{E}\,t$. From the condition $x_\text{A} = x_\text{E}$ it follows that $x_{0\text{A}} + v_\text{A}\,t = x_{0\text{E}} + v_\text{E}\,t$, so $(v_\text{A} - v_\text{E})\,t = x_{0\text{E}} - x_{0\text{A}}$ and

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

   The position is $x = x_{0\text{A}} + v_\text{A}\,t$

   Conversions

   All quantities are in the units $\text{m}$ and $\text{m/s}$, so no conversions are needed. Eeva's velocity is substituted as negative, because she walks in the negative direction.

   Substitution

   $$\begin{aligned} t &= \dfrac{90\,\text{m} - 0}{1.5\,\text{m/s} - (-1.2\,\text{m/s})} \\ &= \dfrac{90\,\text{m}}{2.7\,\text{m/s}} \\ &= 33.33\ldots\,\text{s} \end{aligned}$$

   $$\begin{aligned} x &= 0 + 1.5\,\text{m/s} \cdot 33.33\ldots\,\text{s} \\ &= 50\,\text{m} \end{aligned}$$

   Answer

   $50\,\text{m}$

   Or $5.0 \cdot 10^1\,\text{m}$, two significant figures, because the velocities $1.5\,\text{m/s}$ and $-1.2\,\text{m/s}$ have two figures and so does their difference $2.7\,\text{m/s}$.

   Reasonableness

   Eeva walks $1.2\,\text{m/s} \cdot 33.33\,\text{s} = 40\,\text{m}$, so her position is $90\,\text{m} - 40\,\text{m} = 50\,\text{m}$, the same as Aino's. The position is between the starting and finishing points, and because Aino is faster, the meeting is closer to Eeva's starting position than halfway ($45\,\text{m}$).

3. Exercise 1-18

   The $(t, x)$ graph of a robot consists of three straight parts: between 0 and $10\,\text{s}$ the position grows from 0 to $30\,\text{m}$, between 10 and $20\,\text{s}$ the robot is at rest and between 20 and $40\,\text{s}$ the position decreases to $10\,\text{m}$.

   What is the average velocity of the robot over the whole $40\,\text{s}$? Give the answer with its sign.

   m/sCheck
   Answer

   $0.25\,\text{m/s}$

   Show the solution

   Given

   parts of the $(t, x)$ graph: $x = 0$ at $0\,\text{s}$, $x = 30\,\text{m}$ at $10\,\text{s}$, $x = 30\,\text{m}$ at $20\,\text{s}$ and $x = 10\,\text{m}$ at $40\,\text{s}$

   Find

   the average velocity $v_\text{avg}$ over the whole $40\,\text{s}$ with its sign in the unit $\text{m/s}$

   Principle

   The average velocity is the whole displacement divided by the whole time. Only the initial and final positions affect the displacement, so the intermediate stages, that is moving forward, stopping and coming back, do not affect the calculation.

   Reading the graph

   From the given points we read the starting point $(0\,\text{s}$; $0\,\text{m})$ and the end point $(40\,\text{s}$; $10\,\text{m})$. Between $10\text{–}20\,\text{s}$ the graph is horizontal, because the robot stands still, and its points are not needed. The average velocity is the slope of the line between these two points.

   Formula

   $v_\text{avg}$$= \dfrac{\Delta x}{\Delta t}$

   $= \dfrac{x_2 - x_1}{t_2 - t_1}$

   Conversions

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

   Substitution

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

   $= 0.25\,\text{m/s}$

   Answer

   $0.25\,\text{m/s}$

   Two significant figures, because the final position $10\,\text{m}$ and the time $40\,\text{s}$ are given with two figures.

   Reasonableness

   The robot went $30\,\text{m}$ forward and $20\,\text{m}$ back, so the displacement is $30\,\text{m} - 20\,\text{m} = 10\,\text{m}$ even though the distance was $50\,\text{m}$. The sign is positive, because at the end the robot is at a greater $x$ than at the beginning. The mean of the partial velocities $3.0\,\text{m/s}$, $0$ and $-1.0\,\text{m/s}$,

   $\dfrac{3.0 + 0 - 1.0}{3}$$= 0.67\,\text{m/s}$

   would be wrong, because the sub-intervals are not of equal length.

4. Exercise 1-19

   The motion of a cart was studied with video analysis. The scale of the program was set with a $1.00\,\text{m}$ measuring stick placed beside the track, but the stick was closer to the camera than the track of the cart. The table gives the positions given by the program (example data). The motion of the cart is uniform.

   Later a $2.00\,\text{m}$ measuring tape was placed on the track, and the program gave its length as $1.60\,\text{m}$.

   Determine from the table the velocity given by the program by fitting a straight line. Correct the velocity with the measuring tape to the true velocity $v$ of the cart. Estimate when the cart would be at the true position $6.0\,\text{m}$, and justify whether the prediction is reliable. Also explain why the scale error is systematic and is not corrected by repeating the measurement.

   Write the true velocity of the cart in the answer box.

   Position of the cart on the scale of the program (example data)

   $t\,(\text{s})$	$x_\text{prog}\,(\text{m})$
   0.0	0.30
   1.0	0.95
   2.0	1.62
   3.0	2.21
   4.0	2.88
   5.0	3.52

   m/sCheck
   Answer

   $0.80\,\text{m/s}$

   Show the solution

   Given

   the positions of the program $x_\text{prog}$ at the moments $0.0\ldots5.0\,\text{s}$ (e.g. $(0.0\,\text{s}$; $0.30\,\text{m})$ and $(5.0\,\text{s}$; $3.52\,\text{m})$); the true length of the measuring tape $2.00\,\text{m}$, the length given by the program $1.60\,\text{m}$; the motion is uniform

   Find

   the true velocity $v$ of the cart in the unit $\text{m/s}$, the moment when the true position is $6.0\,\text{m}$, and the reliability of the prediction

   Principle

   In uniform motion the points lie on a line, and the velocity is the slope of the line. A scale error multiplies all lengths given by the program by the same number. It is found by comparing the true value of a known length with the value of the program, and the correction multiplies the velocity by the same number.

   Reading the graph

   The points are marked in the $(t, x_\text{prog})$ graph and a line is fitted through them. The line passes close to the points $(0.0\,\text{s}$; $0.31\,\text{m})$ and $(5.0\,\text{s}$; $3.52\,\text{m})$. With them the velocity given by the program is

   $v_\text{prog}$$= \dfrac{3.52\,\text{m} - 0.31\,\text{m}}{5.0\,\text{s}}$

   $\approx 0.64\,\text{m/s}$

   and the initial position of the program is $0.31\,\text{m}$.

   Formula

   correction factor

   $k$$= \dfrac{\ell_\text{true}}{\ell_\text{prog}}$

   true velocity $v = k\,v_\text{prog}$ and position $x = k\,(x_{0,\text{prog}} + v_\text{prog}\,t)$

   Conversions

   All lengths are in metres and the times in seconds, so no unit conversions are needed.

   Substitution

   $$\begin{aligned} k &= \dfrac{2.00\,\text{m}}{1.60\,\text{m}} \\ &= 1.25 \end{aligned}$$

   $$\begin{aligned} v &= 1.25 \cdot 0.64\,\text{m/s} \\ &= 0.80\,\text{m/s} \end{aligned}$$

   The true initial position is $1.25 \cdot 0.31\,\text{m} = 0.39\,\text{m}$, so from the condition $x = 6.0\,\text{m}$ we get

   $$\begin{aligned} t &= \dfrac{6.0\,\text{m} - 0.39\,\text{m}}{0.80\,\text{m/s}} \\ &= 7.0\,\text{s} \end{aligned}$$

   Answer

   the true velocity is $0.80\,\text{m/s}$, two significant figures like the values of the measuring tape. The $0.64\,\text{m/s}$ given by the program is $20\,\%$ too small. The cart is at the true position $6.0\,\text{m}$ after about $7.0\,\text{s}$.

   Reasonableness

   The measuring tape looked shorter than it is, so all lengths of the program are too small and the correction factor must be greater than one, as $1.25$ is. Checking backwards: $0.80\,\text{m/s} \cdot 5.0\,\text{s} = 4.0\,\text{m}$ is the true displacement of the cart, and the displacement of the program, $3.2\,\text{m}$, is $0.8$ times that. The prediction $7.0\,\text{s}$ is beyond the end of the measurement interval: the measurement ends at the moment $5.0\,\text{s}$, so the prediction extends about $40\,\%$ of the measuring time further. It is reliable only if the track is level and the cart continues at a constant velocity and the track does not end before $6.0\,\text{m}$. The scale error is systematic: it distorts every measurement in the same direction by the same factor, so repeating the measurements does not remove it in the mean. A random error, on the other hand, scatters the points on both sides of the line and decreases with repetition. The error is corrected only by calibrating the scale with a known length from the same distance as the track of the cart.

5. Exercise 1-20

   On a highway, a sensor beside the road follows the position of a truck. The table gives the position of the truck on an $x$-axis along the road (example data). The motion of the truck is uniform and in the positive direction.

   A police car passes the position $x = 0$ at the moment $t = 8.0\,\text{s}$ and drives in the direction of the truck at a uniform velocity $16\,\text{m/s}$.

   Fit a straight line to the measurement points of the truck and write the position of the truck as a function of time. Write the position of the police car as a function of time. Determine at what position the police car catches up with the truck, and justify why this prediction is uncertain when the data of the sensor ends at the moment $t = 40\,\text{s}$.

   Write the position where the police car catches up in the answer box.

   Position of the truck (example data)

   $t\,(\text{s})$	$x\,(\text{m})$
   0	205
   10	302
   20	395
   30	504
   40	598

   mCheck
   Answer

   $7.4 \cdot 10^2\,\text{m}$

   Show the solution

   Given

   the points $(t$; $x)$ of the truck at the moments $0\ldots40\,\text{s}$, e.g. $(0\,\text{s}$; $205\,\text{m})$ and $(40\,\text{s}$; $598\,\text{m})$; police car: $x = 0$ at the moment $t_0 = 8.0\,\text{s}$, velocity $v_\text{P} = 16\,\text{m/s}$

   Find

   the position $x$ where the police car catches up with the truck, in the unit $\text{m}$, and the reliability of the prediction

   Principle

   The motion of the truck is uniform, so its $(t, x)$ points are fitted with a line whose slope is the velocity and whose intercept is the initial position. The position of the police car is the equation of uniform motion, where the time is counted from its moment of passing. The car catches up with the truck when the positions are equal at the same moment.

   Reading the graph

   The points are marked in the graph and a line is fitted through them. The line passes close to the points $(0\,\text{s}$; $203\,\text{m})$ and $(40\,\text{s}$; $598\,\text{m})$. From them the velocity of the truck is

   $v_\text{T}$$= \dfrac{598\,\text{m} - 203\,\text{m}}{40\,\text{s}}$

   $\approx 9.9\,\text{m/s}$

   and the initial position $x_{0\text{T}} \approx 203\,\text{m}$.

   Formula

   truck $x_\text{T} = x_{0\text{T}} + v_\text{T}\,t$ and police car $x_\text{P} = v_\text{P}\,(t - t_0)$. From the condition $x_\text{T} = x_\text{P}$ we get
   $$t = \dfrac{x_{0\text{T}} + v_\text{P}\,t_0}{v_\text{P} - v_\text{T}}$$
   and the position is $x = x_{0\text{T}} + v_\text{T}\,t$.

   Conversions

   All quantities are in the units $\text{m}$, $\text{s}$ and $\text{m/s}$, so no conversions are needed.

   Substitution

   $$\begin{aligned} t &= \dfrac{203\,\text{m} + 16\,\text{m/s} \cdot 8.0\,\text{s}}{16\,\text{m/s} - 9.9\,\text{m/s}} \\ &= \dfrac{331\,\text{m}}{6.1\,\text{m/s}} \\ &= 54\,\text{s} \end{aligned}$$

   $$\begin{aligned} x &= 203\,\text{m} + 9.9\,\text{m/s} \cdot 54\,\text{s} \\ &= 7.4 \cdot 10^2\,\text{m} \end{aligned}$$

   Answer

   $7.4 \cdot 10^2\,\text{m}$

   Or about $740\,\text{m}$, two significant figures, because the difference of the velocities, $16 - 9.9 = 6.1\,\text{m/s}$, limits the accuracy.

   Reasonableness

   The position of the police car at the moment $54\,\text{s}$ is

   $16\,\text{m/s} \cdot (54\,\text{s} - 8.0\,\text{s})$

   $= 7.4 \cdot 10^2\,\text{m}$

   the same as the position of the truck, so the solution checks out. The police car is faster than the truck, so catching up is possible. The prediction is uncertain, because the data ends at the moment $40\,\text{s}$ and the catching up happens at the moment $54\,\text{s}$, that is $14\,\text{s}$ outside the data. The prediction assumes that the truck keeps driving at the same velocity along the same straight road all the time, which the data does not prove. A small error in the velocity of the truck grows in the prediction, because the difference of the velocities is small: if the velocity of the truck were $0.2\,\text{m/s}$ greater, the time would be about $3\,\%$ longer.