Acceleration and uniformly accelerated motion

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

1. Exercise 2-1Core

   A car starts from rest at a traffic light and reaches a speed of $25\,\text{m/s}$ in $8.0\,\text{s}$. The acceleration is uniform. What is the car's acceleration?

   m/s²Check
   Answer

   $3.1\,\text{m/s}^2$

   Show the solution

   Given

   initial velocity $v_0 = 0$ (from rest), final velocity $v = 25\,\text{m/s}$, time $\Delta t = 8.0\,\text{s}$; the acceleration is uniform

   Find

   the car's acceleration $a$ in $\text{m/s}^2$

   Principle

   Acceleration is the change in velocity divided by the time it took. The acceleration is uniform, so the velocity grows by the same amount every second and the average over the whole interval equals the acceleration at every moment.

   Formula

   $a = \dfrac{\Delta v}{\Delta t}$, where $\Delta v = v - v_0$. So $a = \dfrac{v - v_0}{\Delta t}$

   Conversions

   The velocity is in $\text{m/s}$ and the time in seconds, so the result comes out directly in $\text{m/s}^2$. No conversions are needed.

   Substitution

   $a$$= \dfrac{25\,\text{m/s} - 0\,\text{m/s}}{8.0\,\text{s}}$

   $= 3.125\,\text{m/s}^2$

   Answer

   $3.1\,\text{m/s}^2$

   Two significant figures like the velocity $25\,\text{m/s}$ and the time $8.0\,\text{s}$.

   Reasonableness

   The velocity grows by about $3\,\text{m/s}$ every second, and $8 \cdot 3\,\text{m/s} = 24\,\text{m/s}$, which is close to the given $25\,\text{m/s}$. Reaching $25\,\text{m/s} = 90\,\text{km/h}$ in eight seconds is a brisk acceleration for an ordinary passenger car. The acceleration is positive because the velocity grows.

2. Exercise 2-2Core

   A cyclist is riding at $6.0\,\text{m/s}$ and starts to brake. During the braking the acceleration is $-1.5\,\text{m/s}^2$. What is the cyclist's velocity $3.0\,\text{s}$ after the braking starts?

   m/sCheck
   Answer

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

   Show the solution

   Given

   initial velocity $v_0 = 6.0\,\text{m/s}$, acceleration $a = -1.5\,\text{m/s}^2$, time $t = 3.0\,\text{s}$; the direction of motion is positive

   Find

   the velocity $v$ at $t = 3.0\,\text{s}$ in $\text{m/s}$

   Principle

   The acceleration of braking is constant and opposite to the velocity, so the velocity decreases by $1.5\,\text{m/s}$ every second. In uniformly accelerated motion the velocity is the initial velocity plus the product of the acceleration and the time.

   Formula

   $v = v_0 + a\,t$. The formula is already solved for the velocity.

   Conversions

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

   Substitution

   $v$$= 6.0\,\text{m/s} + (-1.5\,\text{m/s}^2) \cdot 3.0\,\text{s}$

   $= 6.0\,\text{m/s} - 4.5\,\text{m/s}$

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

   Answer

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

   Two significant figures like the given values. The difference in the calculation is known to the nearest tenth.

   Reasonableness

   The cyclist would stop in the time $6.0\,\text{m/s} : 1.5\,\text{m/s}^2 = 4.0\,\text{s}$, so at $3.0\,\text{s}$ she is still moving slowly forward. The velocity is positive and smaller than at the start, as it should be in braking.

3. Exercise 2-3

   A stone is dropped from rest into a well. Air resistance is small. How far does the stone fall in $1.50\,\text{s}$?

   mCheck
   Answer

   $11.0\,\text{m}$

   Show the solution

   Given

   initial velocity $v_0 = 0$, time $t = 1.50\,\text{s}$, acceleration of free fall from the table $g = 9.81\,\text{m/s}^2$; air resistance is small

   Find

   the distance fallen $s$ in metres

   Principle

   Air resistance is small, so the stone falls with the constant acceleration $g$. Take downwards as the positive direction. The stone starts from rest, so the displacement is the area of the triangle in the $(t,v)$ graph, $\tfrac{1}{2}\,g\,t^2$.

   Formula

   $x = x_0 + v_0\,t + \tfrac{1}{2}\,a\,t^2$, where $x - x_0 = s$, $v_0 = 0$ and $a = g$. So $s = \dfrac{g\,t^2}{2}$

   Conversions

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

   Substitution

   $s$$= \tfrac{1}{2} \cdot 9.81\,\text{m/s}^2 \cdot (1.50\,\text{s})^2$

   $= \tfrac{1}{2} \cdot 9.81\,\text{m/s}^2 \cdot 2.25\,\text{s}^2$

   $\approx 11.04\,\text{m}$

   Answer

   $11.0\,\text{m}$

   Three significant figures like the time $1.50\,\text{s}$.

   Reasonableness

   The final velocity is $9.81\,\text{m/s}^2 \cdot 1.50\,\text{s} \approx 14.7\,\text{m/s}$, so the average velocity is about $7.4\,\text{m/s}$ and $7.4\,\text{m/s} \cdot 1.50\,\text{s} \approx 11\,\text{m}$. In the first second the stone falls about $5\,\text{m}$, so a little over $10\,\text{m}$ in one and a half seconds is plausible.

4. Exercise 2-4

   The graph shows the velocity of a kick scooter as a function of time. What is the scooter's acceleration?

   m/s²Check
   Answer

   $1.0\,\text{m/s}^2$

   Show the solution

   Given

   from the graph, the velocity $v_1 = 2.0\,\text{m/s}$ at $t_1 = 0$ and $v_2 = 7.0\,\text{m/s}$ at $t_2 = 5.0\,\text{s}$; the graph is a straight line

   Find

   the scooter's acceleration $a$ in $\text{m/s}^2$

   Principle

   The $(t,v)$ graph is a straight line, so the acceleration is constant and equals the slope of the line, that is, the change in velocity divided by the change in time.

   Reading the graph

   The end points of the line are at crossings of grid lines: the starting point $(0$; $2.0\,\text{m/s})$ and the end point $(5.0\,\text{s}$; $7.0\,\text{m/s})$. Points far apart give the most accurate slope.

   Formula

   $a$$= \dfrac{\Delta v}{\Delta t}$

   $= \dfrac{v_2 - v_1}{t_2 - t_1}$

   Conversions

   The velocities are in $\text{m/s}$ and the times in seconds, so the result comes out in $\text{m/s}^2$. No conversions are needed.

   Substitution

   $a$$= \dfrac{7.0\,\text{m/s} - 2.0\,\text{m/s}}{5.0\,\text{s} - 0\,\text{s}}$

   $= 1.0\,\text{m/s}^2$

   Answer

   $1.0\,\text{m/s}^2$

   Two significant figures like the values read from the graph.

   Reasonableness

   In the graph the velocity grows by one $\text{m/s}$ square for every one-second square, so the acceleration $1\,\text{m/s}^2$ can be seen straight from the grid. The acceleration is positive because the line rises.

5. Exercise 2-5

   A train starts from rest at a station and accelerates uniformly with acceleration $0.40\,\text{m/s}^2$. How far does the train travel in the first $30\,\text{s}$?

   mCheck
   Answer

   $180\,\text{m}$

   Show the solution

   Given

   initial velocity $v_0 = 0$, acceleration $a = 0.40\,\text{m/s}^2$, time $t = 30\,\text{s}$

   Find

   the distance travelled $\Delta x$ in metres

   Principle

   The train accelerates uniformly from rest and does not turn round, so the distance is the displacement $\Delta x = v_0\,t + \tfrac{1}{2}\,a\,t^2$, in which the term of the initial velocity is zero.

   Formula

   $x = x_0 + v_0\,t + \tfrac{1}{2}\,a\,t^2$, so $v_0 = 0$ gives

   $\Delta x$$= x - x_0$

   $= \tfrac{1}{2}\,a\,t^2$

   Conversions

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

   Substitution

   $\Delta x$$= \tfrac{1}{2} \cdot 0.40\,\text{m/s}^2 \cdot (30\,\text{s})^2$

   $= \tfrac{1}{2} \cdot 0.40\,\text{m/s}^2 \cdot 900\,\text{s}^2$

   $= 180\,\text{m}$

   Answer

   $180\,\text{m}$

   Or $1.8 \cdot 10^2\,\text{m}$, two significant figures like the acceleration $0.40\,\text{m/s}^2$ and the time $30\,\text{s}$.

   Reasonableness

   The final velocity is $0.40\,\text{m/s}^2 \cdot 30\,\text{s} = 12\,\text{m/s}$, so the average velocity is $6.0\,\text{m/s}$ and $6.0\,\text{m/s} \cdot 30\,\text{s} = 180\,\text{m}$. The final velocity

   $12\,\text{m/s}$$= 12 \cdot 3.6\,\text{km/h}$

   $\approx 43\,\text{km/h}$

   is an ordinary speed for a train leaving a station, and $180\,\text{m}$ is about two lengths of a sprint track ($100\,\text{m}$), roughly the length of a station platform.

6. Exercise 2-6

   A car drives east and brakes uniformly to a stop. In which direction does the car's acceleration point during the braking?

   Choose an option.

   • Upwards, because the car being stopped is acted on by the reaction to its weight, which points upwards.
   • West, because the speed decreases and the acceleration points in the direction of the change in velocity.
   • East, because the car moves east and the acceleration always points in the direction of motion.
   • Nowhere, because the car slows down and the speed does not increase, so the acceleration is zero.

   Answer
   I have done my solution.
   Answer

   West, because the speed decreases and the acceleration points in the direction of the change in velocity.

   Show the solution

   Find

   in which direction the acceleration of a car driving east and braking points.

   Principle

   Acceleration is the change in velocity divided by time, $a = \Delta v : \Delta t$, so it points in the direction of the change in velocity. In braking the speed decreases: the final velocity is smaller than the initial velocity, and the change $\Delta v = v_2 - v_1$ points in the opposite direction to the motion. When the motion is eastwards, the acceleration points west. On a straight line this shows as a sign: if east is positive, the acceleration is negative.

   Model answer

   The acceleration points west, because the speed decreases and the acceleration points in the direction of the change in velocity. It does not point in the direction of motion, because then the car would speed up. It is not zero, because the velocity changes, and it has no vertical component, because the velocity changes only in the horizontal direction.

7. Exercise 2-7

   The graph shows the velocity of a kick scooter as a function of time from $0$ to $5.0\,\text{s}$. The scooter moves in the same direction the whole time.

   How far does the scooter travel during this time?

   mCheck
   Answer

   $22.5\,\text{m}$

   Show the solution

   Given

   from the graph the velocity $v_1 = 2.0\,\text{m/s}$ at $t = 0$ and $v_2 = 7.0\,\text{m/s}$ at $t = 5.0\,\text{s}$; the graph is a straight line and the velocity stays positive

   Find

   the distance travelled $\Delta x$ in metres

   Principle

   The scooter moves in the same direction the whole time, so the distance is the area between the $(t,v)$ graph and the time axis. Under the graph there is a rectangle (height $v_1$) with a triangle (height $v_2 - v_1$) on top of it, that is, a trapezoid.

   Reading the graph

   The endpoints of the line are at intersections of grid lines: the starting point $(0$; $2.0\,\text{m/s})$ and the end point $(5.0\,\text{s}$; $7.0\,\text{m/s})$.

   Formula

   $\Delta x = v_1\,t + \tfrac{1}{2}\,(v_2 - v_1)\,t$, that is, the average of the velocities times the time

   Conversions

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

   Substitution

   $\Delta x$$= 2.0\,\text{m/s} \cdot 5.0\,\text{s} + \tfrac{1}{2} \cdot 5.0\,\text{m/s} \cdot 5.0\,\text{s}$

   $= 10\,\text{m} + 12.5\,\text{m}$

   $= 22.5\,\text{m}$

   Answer

   $22.5\,\text{m}$

   Three significant figures, because the values read are exactly on grid lines.

   Reasonableness

   The average velocity is $(2.0\,\text{m/s} + 7.0\,\text{m/s}) : 2 = 4.5\,\text{m/s}$, and $4.5\,\text{m/s} \cdot 5.0\,\text{s} = 22.5\,\text{m}$. The distance is greater than the $10\,\text{m}$ travelled at the initial velocity and smaller than the $35\,\text{m}$ travelled at the final velocity, as it should be.

8. Exercise 2-8

   A student solves a problem: a stone is dropped from rest into a well and it reaches the bottom in $2.0\,\text{s}$. How deep is the well?

   The student's solution: the velocity at $2.0\,\text{s}$ is $v = g\,t = 9.81\,\text{m/s}^2 \cdot 2.0\,\text{s} \approx 19.6\,\text{m/s}$. The depth is $s = v\,t = 19.6\,\text{m/s} \cdot 2.0\,\text{s} \approx 39\,\text{m}$.

   What is wrong in the student's solution?

   Choose an option.

   • The stone's velocity increases the whole time, so the distance must not be calculated with the final velocity but with the average velocity.
   • The fall time must be divided by the velocity, because the distance is obtained from the formula $s = t : v$.
   • The acceleration of free fall must be multiplied by the stone's mass, because a heavy stone falls faster than a light one.
   • The stone's acceleration is not constant, so the formula $v = g\,t$ must not be used to calculate the velocity.

   Answer
   I have done my solution.
   Answer

   The stone's velocity increases the whole time, so the distance must not be calculated with the final velocity but with the average velocity.

   Show the solution

   Find

   what is wrong in the student's solution.

   Principle

   The stone falls with constant acceleration $g$, so the velocity $v = g\,t$ grows uniformly from zero to $19.6\,\text{m/s}$. The distance $v\,t$ holds only if the velocity is the same the whole time. Here the velocity changes the whole time, so the distance is the area of the triangle under the $(t,v)$ graph, that is, the average velocity times the time.

   Model answer

   The error is that the distance was calculated with the final velocity, although the stone's velocity increases during the whole fall. The calculation of the velocity was correct. The correct distance comes from the average velocity: $\tfrac{1}{2} \cdot 19.6\,\text{m/s} \cdot 2.0\,\text{s} \approx 20\,\text{m}$, which is half of the student's result. The depth of the well is therefore about $20\,\text{m}$. The mass is not needed, because the acceleration of free fall is the same for all bodies when air resistance is small.

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

1. Exercise 2-9Core

   A car is driving at $72\,\text{km/h}$ and brakes uniformly. It stops in $4.0\,\text{s}$. How far does the car travel during the braking?

   mCheck
   Answer

   $40\,\text{m}$

   Show the solution

   Given

   initial velocity $v_0 = 72\,\text{km/h}$, final velocity $v = 0$, braking time $t = 4.0\,\text{s}$; the braking is uniform

   Find

   the braking distance $\Delta x$ in metres

   Principle

   The braking is uniform, so the acceleration is constant. The acceleration comes from the change in velocity and the time, and then the distance comes from the equation of position. The car does not turn round, so the distance equals the displacement.

   Formula

   $a = \dfrac{v - v_0}{t}$ and $\Delta x = v_0\,t + \tfrac{1}{2}\,a\,t^2$

   Conversions

   The velocity is converted by substituting units:

   $72\,\text{km/h}$$= 72 \cdot \tfrac{1000\,\text{m}}{3600\,\text{s}}$

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

   Substitution

   $a$$= \dfrac{0 - 20\,\text{m/s}}{4.0\,\text{s}}$

   $= -5.0\,\text{m/s}^2$

   $\Delta x$$= 20\,\text{m/s} \cdot 4.0\,\text{s} + \tfrac{1}{2} \cdot (-5.0\,\text{m/s}^2) \cdot (4.0\,\text{s})^2$

   $= 80\,\text{m} - 40\,\text{m}$

   $= 40\,\text{m}$

   Answer

   $40\,\text{m}$

   Two significant figures like the velocity $72\,\text{km/h}$ and the time $4.0\,\text{s}$.

   Reasonableness

   The velocity decreases steadily from $20\,\text{m/s}$ to zero, so the average velocity is $10\,\text{m/s}$ and $10\,\text{m/s} \cdot 4.0\,\text{s} = 40\,\text{m}$, the same result. The magnitude of the acceleration, $5\,\text{m/s}^2$, is about half of $g$, which is an ordinary hard braking on a dry road.

2. Exercise 2-10Core

   The graph shows the position of a cart accelerating from rest as a function of time. A tangent has been drawn as a dashed line at the point $P$ of the curve at $4.0\,\text{s}$.

   What is the cart's instantaneous velocity at $4.0\,\text{s}$?

   m/sCheck
   Answer

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

   Show the solution

   Given

   from the graph, the points of the tangent $(t_1$; $x_1) = (2.0\,\text{s}$; $0\,\text{m})$ and $(t_2$; $x_2) = (6.0\,\text{s}$; $16\,\text{m})$; the tangent is drawn at $t = 4.0\,\text{s}$

   Find

   the instantaneous velocity $v$ at $4.0\,\text{s}$ in $\text{m/s}$

   Principle

   The $(t,x)$ graph is a curve because the velocity changes. The velocity at a given moment is the slope of the tangent of the curve at that point. The slope is calculated from the tangent, not from points of the curve.

   Reading the graph

   The tangent crosses the time axis at $t = 2.0\,\text{s}$, where $x = 0$, and passes through the grid crossing $(6.0\,\text{s}$; $16\,\text{m})$. These points are far apart and are read exactly.

   Formula

   $v$$= \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 the result comes out in $\text{m/s}$. No conversions are needed.

   Substitution

   $v$$= \dfrac{16\,\text{m} - 0\,\text{m}}{6.0\,\text{s} - 2.0\,\text{s}}$

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

   Answer

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

   Two significant figures like the values read from the graph.

   Reasonableness

   In the graph the cart is at the position $8.0\,\text{m}$ at $4.0\,\text{s}$, so its average velocity from the start is $2.0\,\text{m/s}$. For an object accelerating uniformly from rest the final velocity is twice the average velocity, that is, $4.0\,\text{m/s}$, the same as from the tangent.

3. Exercise 2-11

   A ball is thrown straight up at $15.0\,\text{m/s}$. Air resistance is small. How high above the point of release does the ball rise?

   mCheck
   Answer

   $11.5\,\text{m}$

   Show the solution

   Given

   initial velocity $v_0 = 15.0\,\text{m/s}$ upwards, acceleration of free fall from the table $g = 9.81\,\text{m/s}^2$ downwards; air resistance is small

   Find

   the height of the rise $h$ from the point of release in metres

   Principle

   Take upwards as the positive direction, so the acceleration is $-g$ during the whole flight. At the highest point the velocity is zero for a moment. First solve the time of the rise from the condition $v = 0$, then the height from the equation of position.

   Formula

   $v = v_0 - g\,t = 0$, so $t = \dfrac{v_0}{g}$, and the height is $h = v_0\,t - \dfrac{g\,t^2}{2}$

   Conversions

   All quantities are in SI units. No conversions are needed.

   Substitution

   $t$$= \dfrac{15.0\,\text{m/s}}{9.81\,\text{m/s}^2}$

   $\approx 1.529\,\text{s}$

   $h$$= 15.0\,\text{m/s} \cdot 1.529\,\text{s} - \tfrac{1}{2} \cdot 9.81\,\text{m/s}^2 \cdot (1.529\,\text{s})^2$

   $\approx 22.94\,\text{m} - 11.47\,\text{m}$

   $\approx 11.47\,\text{m}$

   Answer

   $11.5\,\text{m}$

   Three significant figures like the initial velocity $15.0\,\text{m/s}$.

   Reasonableness

   During the rise the velocity decreases steadily from $15.0\,\text{m/s}$ to zero, so the average velocity is $7.5\,\text{m/s}$ and $7.5\,\text{m/s} \cdot 1.529\,\text{s} \approx 11.5\,\text{m}$. A height of about three storeys is plausible for a hard throw.

4. Exercise 2-12

   On the Moon an astronaut drops a hammer from rest from a height of $1.60\,\text{m}$, and the hammer hits the surface $1.41\,\text{s}$ later. The Moon has no atmosphere.

   What is the acceleration of free fall on the Moon according to this measurement?

   m/s²Check
   Answer

   $1.61\,\text{m/s}^2$

   Show the solution

   Given

   drop height $h = 1.60\,\text{m}$, fall time $t = 1.41\,\text{s}$, initial velocity $v_0 = 0$; there is no air resistance on the Moon

   Find

   the Moon's acceleration of free fall $g_\text{M}$ in $\text{m/s}^2$

   Principle

   There is no air resistance, so the hammer falls with constant acceleration. It starts from rest, so the drop height is $\tfrac{1}{2}\,g_\text{M}\,t^2$. Solve the acceleration from this.

   Formula

   $h = \tfrac{1}{2}\,g_\text{M}\,t^2$, so $g_\text{M} = \dfrac{2h}{t^2}$

   Conversions

   The height is in metres and the time in seconds, so the result comes out in $\text{m/s}^2$. No conversions are needed.

   Substitution

   $g_\text{M}$$= \dfrac{2 \cdot 1.60\,\text{m}}{(1.41\,\text{s})^2}$

   $= \dfrac{3.20\,\text{m}}{1.9881\,\text{s}^2}$

   $\approx 1.610\,\text{m/s}^2$

   Answer

   $1.61\,\text{m/s}^2$

   Three significant figures like the height $1.60\,\text{m}$ and the time $1.41\,\text{s}$.

   Reasonableness

   The table value for the Moon's acceleration of free fall is $1.62\,\text{m/s}^2$, about one sixth of the Earth's value, so the result is close to it. On Earth the same drop would take only $\sqrt{2 \cdot 1.60\,\text{m} : 9.81\,\text{m/s}^2} \approx 0.57\,\text{s}$; the slower fall on the Moon corresponds to the smaller acceleration.

5. Exercise 2-13

   The velocity of a cart rolling down an inclined plane was measured as a function of time by video analysis. The data is an example data set.

   Plot the measured points as a $(t,v)$ graph and fit a straight line to the points. Determine the cart's acceleration from the slope of the line. Give the answer in $\text{m/s}^2$.

   Velocity of the cart, example data

   $t\,(\text{s})$	$v\,(\text{m/s})$
   0.0	0.21
   0.5	0.66
   1.0	1.02
   1.5	1.50
   2.0	1.86
   2.5	2.35
   3.0	2.71

   m/s²Check
   Answer

   $0.84\,\text{m/s}^2$

   Show the solution

   Given

   the $(t,v)$ points of the table, seven measurements in the range $0 \ldots 3.0\,\text{s}$; the cart accelerates uniformly

   Find

   the cart's acceleration $a$ in $\text{m/s}^2$

   Principle

   In uniformly accelerated motion the $(t,v)$ graph is a straight line, and its slope is the acceleration, because $v = v_0 + a\,t$. The measured points are not exactly on a line, so the line is fitted to all the points at the same time.

   Reading the graph

   The points $(t,v)$ rise approximately along a straight line. With a calculator or a spreadsheet the equation of the fitted line is $v \approx 0.84\,\text{m/s}^2 \cdot t + 0.22\,\text{m/s}$.

   Formula

   $a = \dfrac{\Delta v}{\Delta t}$, that is, the slope $k$ of the fitted line

   Conversions

   The velocities are in $\text{m/s}$ and the times in seconds, so the slope is in $\text{m/s}^2$. No conversions are needed.

   Substitution

   The slope is obtained from two points of the line far from each other:

   $a$$= \dfrac{2.73\,\text{m/s} - 0.22\,\text{m/s}}{3.0\,\text{s} - 0\,\text{s}}$

   $\approx 0.84\,\text{m/s}^2$

   Check with the end points of the data:

   $\dfrac{2.71\,\text{m/s} - 0.21\,\text{m/s}}{3.0\,\text{s}}$

   $\approx 0.83\,\text{m/s}^2$

   Answer

   $0.84\,\text{m/s}^2$

   Two significant figures, because the velocities of the measurements are given to two decimal places and the points scatter slightly.

   Reasonableness

   The initial velocity $0.22\,\text{m/s}$ is close to the measured $0.21\,\text{m/s}$, and the value of the fit at $3.0\,\text{s}$ is

   $0.22\,\text{m/s} + 0.84\,\text{m/s}^2 \cdot 3.0\,\text{s}$

   $\approx 2.74\,\text{m/s}$

   close to the measured $2.71\,\text{m/s}$. The acceleration is small compared with $g = 9.81\,\text{m/s}^2$, as it should be on a gentle slope.

6. Exercise 2-14

   A cart starts from rest on an inclined track, and its position $x$ from the starting point was measured as a function of time. The data is an example data set.

   The position does not grow in direct proportion to time, so a straight line cannot be fitted to the $(t,x)$ graph. Calculate $t^2$ for each moment, plot the $(t^2,x)$ graph and fit a straight line to the points. Determine the cart's acceleration from the slope of the line. Give the answer in $\text{m/s}^2$.

   Position of the cart, example data

   $t\,(\text{s})$	$x\,(\text{m})$
   0.0	0.00
   0.4	0.10
   0.8	0.37
   1.2	0.80
   1.6	1.46
   2.0	2.32

   m/s²Check
   Answer

   $1.15\,\text{m/s}^2$

   Show the solution

   Given

   the $(t,x)$ points of the table, six measurements in the range $0 \ldots 2.0\,\text{s}$; the cart starts from rest at the position $x_0 = 0$

   Find

   the cart's acceleration $a$ in $\text{m/s}^2$

   Principle

   The cart starts from rest and accelerates uniformly, so $x = \tfrac{1}{2}\,a\,t^2$. Then $x$ is directly proportional to $t^2$: when the $(t^2,x)$ graph is plotted, the points are on a line whose slope is $k = \tfrac{1}{2}\,a$. In this way fitting a curve becomes fitting a line.

   Reading the graph

   Calculate $t^2$ for each moment: $0$, $0.16$, $0.64$, $1.44$, $2.56$ and $4.00$ (unit $\text{s}^2$). The points $(t^2,x)$ fit a line, and the slope of the fitted line is $k \approx 0.577\,\text{m/s}^2$.

   Formula

   $a = 2k$

   Conversions

   The slope is in $\text{m} : \text{s}^2 = \text{m/s}^2$, so no conversions are needed.

   Substitution

   $a$$= 2 \cdot 0.577\,\text{m/s}^2$

   $\approx 1.15\,\text{m/s}^2$

   Check with the end point:

   $k$$\approx 2.32\,\text{m} : 4.00\,\text{s}^2$

   $= 0.58\,\text{m/s}^2$

   so $a \approx 1.16\,\text{m/s}^2$.

   Answer

   $1.15\,\text{m/s}^2$

   Three significant figures like the fitted slope; the scatter of the measurements, however, limits the accuracy to about two significant figures.

   Reasonableness

   According to the fit, the position at $2.0\,\text{s}$ is

   $\tfrac{1}{2} \cdot 1.15\,\text{m/s}^2 \cdot (2.0\,\text{s})^2$

   $= 2.30\,\text{m}$

   close to the measured $2.32\,\text{m}$. The constant term of the line is close to zero, as it should be when the cart starts from the origin.

7. Exercise 2-15

   A cart rolls on an inclined track. A student wants to find out whether the cart's acceleration is constant during the whole trip. She has two ways of measuring. Light gates measure the time between two gates to the nearest millisecond. Video analysis gives the cart's position in every frame, for example thirty times a second.

   Which way of measuring gives a better answer to the question?

   Choose an option.

   • Light gates, because the timing is more accurate and so the acceleration can be deduced from a single measurement.
   • Video analysis, because the velocity is obtained at many moments and we can see whether the $(t,v)$ graph is a straight line.
   • Neither is suitable, because the cart's acceleration can only be determined by calculation from the angle of inclination.
   • Light gates, because the gates measure the velocity directly and from the velocity the acceleration is always constant.

   Answer
   I have done my solution.
   Answer

   Video analysis, because the velocity is obtained at many moments and we can see whether the $(t,v)$ graph is a straight line.

   Show the solution

   Find

   which way of measuring is best for investigating whether the cart's acceleration is constant during the whole trip.

   Principle

   The acceleration is constant if the $(t,v)$ graph is a straight line. To establish this we need many data points from different moments, so that we can see whether the slope stays the same. The accuracy of a measurement is a different matter from the number of data points: even an accurate single time does not tell how the velocity changes along the way.

   Model answer

   A better answer is obtained with video analysis. From each frame we get the cart's position and from it the velocity, so the $(t,v)$ graph gets many points, and we can see whether the graph is a straight line. Two light gates give only one average acceleration, even though the time is measured accurately, so they cannot show whether the acceleration changes. Nor can the acceleration be assumed constant, because that is exactly what is being investigated. Light gates are better suited to a situation where constant acceleration is already known and an accurate single value is wanted.

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

1. Exercise 2-16

   A ball is thrown straight up from a balcony at $8.00\,\text{m/s}$. The point of release is $6.00\,\text{m}$ above the ground, and the ball finally lands on the ground next to the balcony. Air resistance is small.

   How long after the throw does the ball hit the ground?

   sCheck
   Answer

   $2.19\,\text{s}$

   Show the solution

   Given

   initial velocity $v_0 = 8.00\,\text{m/s}$ upwards, point of release $6.00\,\text{m}$ above the ground, acceleration of free fall from the table $g = 9.81\,\text{m/s}^2$; air resistance is small

   Find

   the time $t$ from the throw to hitting the ground, in seconds

   Principle

   The ball's acceleration is $g$ downwards during the whole flight, also at the highest point, so the whole flight is one uniformly accelerated motion. Place the origin at the point of release and take upwards as positive. The ground is then at the position $y = -6.00\,\text{m}$, and the time comes from a quadratic equation.

   Formula

   $y = v_0\,t - \tfrac{1}{2}\,g\,t^2$. When $y = -h$: $\tfrac{1}{2}\,g\,t^2 - v_0\,t - h = 0$, whose positive solution is

   $t$$= \dfrac{v_0 + \sqrt{v_0^2 + 2gh}}{g}$

   Conversions

   All quantities are in SI units. No conversions are needed.

   Substitution

   Under the root

   $v_0^2 + 2gh$$= (8.00\,\text{m/s})^2 + 2 \cdot 9.81\,\text{m/s}^2 \cdot 6.00\,\text{m}$

   $= 181.72\,\text{m}^2/\text{s}^2$

   and $\sqrt{181.72\,\text{m}^2/\text{s}^2} \approx 13.48\,\text{m/s}$. So

   $t$$= \dfrac{8.00\,\text{m/s} + 13.48\,\text{m/s}}{9.81\,\text{m/s}^2}$

   $\approx 2.19\,\text{s}$

   Answer

   $2.19\,\text{s}$

   Three significant figures like the initial velocity and the height. The other solution, $t \approx -0.56\,\text{s}$, is rejected because it is before the throw.

   Reasonableness

   The rise takes $8.00\,\text{m/s} : 9.81\,\text{m/s}^2 \approx 0.82\,\text{s}$ and lifts the ball $3.26\,\text{m}$, so from the highest point to the ground it is $9.26\,\text{m}$. A fall from rest takes $\sqrt{2 \cdot 9.26\,\text{m} : 9.81\,\text{m/s}^2} \approx 1.37\,\text{s}$, and $0.82\,\text{s} + 1.37\,\text{s} = 2.19\,\text{s}$, the same result.

2. Exercise 2-17

   A metro train starts from rest at a station and accelerates uniformly to $15.0\,\text{m/s}$ in $20.0\,\text{s}$. Then it drives at this speed for $60.0\,\text{s}$ and finally brakes uniformly to a stop at the next station in $15.0\,\text{s}$.

   How long is the distance between the stations, in metres?

   mCheck
   Answer

   $1160\,\text{m}$

   Show the solution

   Given

   acceleration $0 \to 15.0\,\text{m/s}$ in $t_1 = 20.0\,\text{s}$, constant speed $v = 15.0\,\text{m/s}$ for $t_2 = 60.0\,\text{s}$, braking to zero in $t_3 = 15.0\,\text{s}$

   Find

   the distance between the stations $s$ in metres

   Principle

   The train always moves in the same direction, so the distance is the area under the $(t,v)$ graph. The graph is a triangle (accelerating), a rectangle (constant speed) and a triangle (braking). The accelerations do not need to be calculated, because the areas come directly from the velocity and the times.

   Formula

   $s = \tfrac{1}{2}\,v\,t_1 + v\,t_2 + \tfrac{1}{2}\,v\,t_3$

   Conversions

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

   Substitution

   $s$$= \tfrac{1}{2} \cdot 15.0\,\text{m/s} \cdot 20.0\,\text{s} + 15.0\,\text{m/s} \cdot 60.0\,\text{s} + \tfrac{1}{2} \cdot 15.0\,\text{m/s} \cdot 15.0\,\text{s}$

   $= 150\,\text{m} + 900\,\text{m} + 112.5\,\text{m}$

   $= 1162.5\,\text{m}$

   Answer

   $1160\,\text{m}$

   Or $1.16\,\text{km}$, three significant figures like the velocity and the times. The result $1162.5\,\text{m}$ is rounded to three significant figures, so the zero at the end of the number is not significant; in kilometres this is clearer.

   Reasonableness

   If the train travelled the whole $95\,\text{s}$ at $15.0\,\text{m/s}$, the distance would be $1425\,\text{m}$; the acceleration and the braking take away half of their own shares, $150\,\text{m} + 112.5\,\text{m}$, so $1425\,\text{m} - 262.5\,\text{m} = 1162.5\,\text{m}$. A distance of a little over a kilometre between stations is usual for a metro.

3. Exercise 2-18

   A ball is thrown straight up, and its height $y$ is measured as a function of time by video analysis. The quadratic polynomial fitted to the measured points is $y = -4.86\,\tfrac{\text{m}}{\text{s}^2} \cdot t^2 + 3.10\,\tfrac{\text{m}}{\text{s}} \cdot t + 1.20\,\text{m}$, with the upward direction positive.

   What is the ball's velocity at $t = 0.500\,\text{s}$? Give the answer with its sign.

   m/sCheck
   Answer

   $-1.76\,\text{m/s}$

   Show the solution

   Given

   the fitted polynomial $y = -4.86\,\tfrac{\text{m}}{\text{s}^2} \cdot t^2 + 3.10\,\tfrac{\text{m}}{\text{s}} \cdot t + 1.20\,\text{m}$, upwards positive; the moment $t = 0.500\,\text{s}$

   Find

   the ball's velocity $v$ with its sign in $\text{m/s}$ at $0.500\,\text{s}$

   Principle

   The polynomial is compared with the equation of position of uniformly accelerated motion, $y = y_0 + v_0\,t + \tfrac{1}{2}\,a\,t^2$. From the coefficients we read the initial velocity $v_0 = 3.10\,\text{m/s}$ and $\tfrac{1}{2}\,a = -4.86\,\text{m/s}^2$, so $a = -9.72\,\text{m/s}^2$. The velocity then comes from the equation $v = v_0 + a\,t$.

   Formula

   From the coefficients of the polynomial $y = A\,t^2 + B\,t + C$ we get $v_0 = B$ and $a = 2A$, so $v = B + 2A\,t$

   Conversions

   The coefficients are in SI units and the time in seconds. No conversions are needed.

   Substitution

   $v$$= 3.10\,\text{m/s} + 2 \cdot (-4.86\,\text{m/s}^2) \cdot 0.500\,\text{s}$

   $= 3.10\,\text{m/s} - 4.86\,\text{m/s}$

   $= -1.76\,\text{m/s}$

   Answer

   $-1.76\,\text{m/s}$

   Three significant figures, because the coefficients of the polynomial and the moment $0.500\,\text{s}$ are given to three significant figures. The negative sign means that the ball is already moving downwards.

   Reasonableness

   The acceleration obtained from the measurement, $-9.72\,\text{m/s}^2$, is close to $-g$, as it should be. The ball is at its highest point at the time $3.10\,\text{m/s} : 9.72\,\text{m/s}^2 \approx 0.32\,\text{s}$, so at $0.500\,\text{s}$ it is already descending and the velocity must be negative. The descent has lasted $0.18\,\text{s}$, and $9.72\,\text{m/s}^2 \cdot 0.18\,\text{s} \approx 1.8\,\text{m/s}$.

4. Exercise 2-19

   The acceleration of a car was investigated on a straight road. The car started from rest, and the speed was read from the speedometer every two seconds. The data is an example data set.

   a) Calculate the car's acceleration in each two-second interval. Decide and justify whether the acceleration is constant.

   b) Calculate the distance the car travels during the first $8.0\,\text{s}$ by reading it from the area under the $(t,v)$ graph. Assume that the velocity changes uniformly in each two-second interval. Give the distance in the answer box in metres and estimate whether the result is slightly too small or slightly too large.

   Velocity of the car, example data

   $t\,(\text{s})$	$v\,(\text{km/h})$
   0.0	0
   2.0	36
   4.0	62
   6.0	83
   8.0	100

   mCheck
   Answer

   $128\,\text{m}$

   Show the solution

   Given

   the car starts from rest; the velocity $v$ at the moments $t = 0$, $2.0$, $4.0$, $6.0$ and $8.0\,\text{s}$ is $0$, $36$, $62$, $83$ and $100\,\text{km/h}$

   Find

   a) the acceleration in each two-second interval and whether it is constant; b) the distance $s$ travelled in $8.0\,\text{s}$ in metres

   Principle

   The car moves in the same direction the whole time. The acceleration in an interval is the change in velocity divided by time, $a = \Delta v : \Delta t$. If the acceleration were constant, the velocity would grow by the same amount in every interval. The distance is the area under the $(t,v)$ graph, and when the velocity is assumed to change uniformly within each interval, the area of each interval is a trapezoid: the average of the velocities times $2.0\,\text{s}$.

   Reading the graph

   The velocities of the table are read as $(t,v)$ points: $(0$; $0)$, $(2.0\,\text{s}$; $36\,\text{km/h})$, $(4.0\,\text{s}$; $62\,\text{km/h})$, $(6.0\,\text{s}$; $83\,\text{km/h})$ and $(8.0\,\text{s}$; $100\,\text{km/h})$. The graph rises but curves towards a gentler slope.

   Formula

   $a$$= \dfrac{v_2 - v_1}{t_2 - t_1}$

   $\Delta s$$= \dfrac{v_1 + v_2}{2} \cdot \Delta t$

   the total distance is the sum of the distances of the intervals

   Conversions

   Convert the velocities to $\text{m/s}$: $v = \dfrac{v_{\text{km/h}}}{3.6}$. The values are $0$, $10.0$, $17.2$, $23.1$ and $27.8\,\text{m/s}$.

   Substitution

   a) The accelerations are ($\text{m/s}^2$) $\dfrac{10.0 - 0}{2.0} = 5.0$, $\dfrac{17.2 - 10.0}{2.0} = 3.6$, $\dfrac{23.1 - 17.2}{2.0} \approx 2.9$ and $\dfrac{27.8 - 23.1}{2.0} \approx 2.4$. b) The distances of the intervals are $10.0\,\text{m}$, $27.2\,\text{m}$, $40.3\,\text{m}$ and $50.8\,\text{m}$. Their sum is $128.3\,\text{m}$.

   Answer

   a) The acceleration is not constant: it decreases in every interval from $5.0\,\text{m/s}^2$ to $2.4\,\text{m/s}^2$, so the $(t,v)$ graph curves towards a gentler slope and is not a straight line. The change is much larger than the inaccuracy of the reading. b) The distance is about $128\,\text{m}$, three significant figures like the conversion of the velocities. The graph is convex upwards, so the straight chord of each interval runs below the curve and the real distance is slightly greater than the calculated one.

   Reasonableness

   If the acceleration were on average $27.8\,\text{m/s} : 8.0\,\text{s} \approx 3.5\,\text{m/s}^2$ the whole time, the distance would be

   $\tfrac{1}{2} \cdot 27.8\,\text{m/s} \cdot 8.0\,\text{s}$

   $\approx 111\,\text{m}$

   The calculated distance is greater, because the car reaches a high speed early. The result is below the $222\,\text{m}$ that would be travelled at the final velocity $27.8\,\text{m/s}$ the whole time, as it should be, and an acceleration of $5\,\text{m/s}^2$ is about half of $g$, which is a usual initial acceleration for a car.

5. Exercise 2-20

   A cart is pushed up an inclined track, and a motion sensor measures its velocity $v$ as a function of time. Up the track is the positive direction, and friction is small. The data is an example data set.

   a) Plot the $(t,v)$ graph and fit a straight line to the points. Determine the cart's acceleration and initial velocity and justify that the acceleration is constant.

   b) How far from the starting point does the cart get upwards before turning? Give the distance in the answer box in metres.

   c) Describe in words what the $(t,a)$ and $(t,x)$ graphs look like from $0$ to $5.0\,\text{s}$.

   Velocity of the cart on the track, example data

   $t\,(\text{s})$	$v\,(\text{m/s})$
   0.0	2.41
   0.5	2.02
   1.0	1.62
   1.5	1.19
   2.0	0.82
   2.5	0.38
   3.0	−0.01
   3.5	−0.41
   4.0	−0.83

   mCheck
   Answer

   $3.6\,\text{m}$

   Show the solution

   Given

   the $(t,v)$ points of the table in the range $0 \ldots 4.0\,\text{s}$, up the track is the positive direction; friction is small

   Find

   a) the acceleration $a$, the initial velocity $v_0$ and the justification for constant acceleration; b) the greatest distance $s$ upwards from the starting point in metres; c) the shapes of the $(t,a)$ and $(t,x)$ graphs

   Principle

   If the $(t,v)$ points are on a line, the acceleration is constant and the motion is uniformly accelerated: $v = v_0 + a\,t$, where the slope is $a$ and the intercept on the vertical axis is $v_0$. The cart turns when $v = 0$. After that it comes back down, because the acceleration still points down the track. The greatest distance is obtained from the equation of position or from the area of the triangle under the $(t,v)$ graph up to the moment $v = 0$.

   Reading the graph

   The points are approximately on a line, and the velocity decreases by about $0.8\,\text{m/s}$ for every second. The fit gives $v \approx -0.81\,\text{m/s}^2 \cdot t + 2.42\,\text{m/s}$, so $a \approx -0.81\,\text{m/s}^2$ and $v_0 \approx 2.42\,\text{m/s}$. The position of the line among the points shows constant acceleration: the points are on both sides of the line without curvature, and the velocity decreases by about the same amount in every $0.5\,\text{s}$ interval ($0.37 \ldots 0.44\,\text{m/s}$).

   Formula

   The turning moment from the condition $v = v_0 + a\,t_k = 0$ is $t_k = -\dfrac{v_0}{a}$, and the distance is $s = v_0\,t_k + \tfrac{1}{2}\,a\,t_k^2$, which simplifies to $s = \dfrac{v_0^2}{2\,|a|}$

   Conversions

   All quantities are in SI units. No conversions are needed.

   Substitution

   $t_k$$= \dfrac{2.42\,\text{m/s}}{0.81\,\text{m/s}^2}$

   $\approx 2.99\,\text{s}$

   The distance is

   $s$$= \dfrac{(2.42\,\text{m/s})^2}{2 \cdot 0.81\,\text{m/s}^2}$

   $= \dfrac{5.86\,\text{m}^2/\text{s}^2}{1.62\,\text{m/s}^2}$

   $\approx 3.61\,\text{m}$

   c) The graphs are not drawn in the answer box but described in words below.

   Answer

   The cart gets about $3.6\,\text{m}$ upwards from the starting point, two significant figures, because the acceleration and the initial velocity have been fitted from scattered data. c) The $(t,a)$ graph is a horizontal line at $a \approx -0.81\,\text{m/s}^2$ for the whole interval, because the acceleration is constant and points down the track also at the turning moment. The $(t,x)$ graph is a downward-opening parabola that starts from the origin, rises steeply at first and flattens, has its peak at $x \approx 3.6\,\text{m}$ at $t \approx 3.0\,\text{s}$ (the tangent is horizontal, because $v = 0$) and falls after that; the cart returns to the starting point at $2\,t_k \approx 6.0\,\text{s}$, so in the range $0 \ldots 5.0\,\text{s}$ the graph is still positive.

   Reasonableness

   The average velocity of the ascent is $2.42\,\text{m/s} : 2 = 1.21\,\text{m/s}$, and $1.21\,\text{m/s} \cdot 2.99\,\text{s} \approx 3.6\,\text{m}$. The last velocity of the data, $-0.83\,\text{m/s}$, is close to the value of the fit, $2.42 - 0.81 \cdot 4.0 \approx -0.82\,\text{m/s}$, and the data includes the turning moment $t \approx 3.0\,\text{s}$ ($v \approx 0$). The acceleration $0.81\,\text{m/s}^2$ is plausible for a gentle slope: it corresponds to an angle of inclination where $g \sin\alpha \approx 0.81\,\text{m/s}^2$, that is, $\alpha \approx 5\,{}^\circ$.