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Section 2.9 One-Dimensional Motion Bootcamp

Subsection 2.9.1 Tracking Motion

Subsection 2.9.2 Position, Displacement, and Distance

Subsection 2.9.3 Speed and Velocity

Subsection 2.9.4 Acceleration

Subsection 2.9.5 Displacement from Velocity

Subsection 2.9.6 Constant Acceleration Motion

Subsection 2.9.7 Freely Falling Motion

Subsection 2.9.8 Miscellaneous

Karl and Justin are in a \(10\text{-km}\) race. Karl runs at a constant speed of \(12\text{ km/h}\) and Justin at \(11.5\text{ km/h}\text{.}\) You can assume \(t = 0 \) at the start and their speeds are their full values at the start.

(a) Who will finish the race first?

(b) What is the value of time, \(t \text{,}\) when Karl finishes?

(c) What is the value of time, \(t \text{,}\) when Justin finishes?

(d) If Karl finishes first, how far away from the finish is Justin when Karl crosses the finish line, and if Justin finishes first, same question for Karl.

(e) How far apart are the two runners after \(12\text{ min} \text{?}\)


(a) Think who is faster? (b) Use \(v = d/t\text{.}\) (c) Same as (b). (d) Use \(d = v t\) on slower with \(t\) the finish time of the faster. (e) Multiply the difference in their speeds by time.


(a) Karl, (b) \(50\text{ min}\text{,}\) (c) \(52.2\text{ min}\text{,}\) (d) \(0.417\text{ km}\text{,}\) (e) \(100\text{ m}\text{.}\)

Solution 1 (a)

(a) Since Karl's speed is greater, he will finish first.

Solution 2 (b)

(b) Using \(v = d/t \) on Karl gives

\begin{equation*} t = \dfrac{d}{v} = \dfrac{10}{12}\text{ hr} = 50\text{ min}. \end{equation*}
Solution 3 (c)

(c) Using \(v = d/t \) on Justin gives

\begin{equation*} t = \dfrac{d}{v} = \dfrac{10}{11.5}\text{ hr} = 52.2\text{ min}. \end{equation*}
Solution 4 (d)

(d) Since Karl finishes at \(t= 50\text{ min} \text{,}\) Justin will have covered the following distance by then (note: I need to convert \(50\text{ min} = 5/6\text{ hr} \) since the speed is given in per \(\text{hr}\) unit.)

\begin{equation*} d = vt = 11.5\times \dfrac{5}{6} = 9.583\text{ km}. \end{equation*}

That means, the remaining distance is

\begin{equation*} 10 - 9.583 = 0.417\text{ km}. \end{equation*}
Solution 5 (e)

(e) Using \(d = v t \) with \(v\) being the difference of the two speeds. Also, since the speeds have time in \(\text{hr}\text{,}\) we need to convert \(t\) to \(\text{hr} \) as well.

\begin{equation*} d = (12-11.5)\times \dfrac{12}{60} = 0.1\text{ km} = 100\text{ m}. \end{equation*}

You can also do this part by first finding how far Karl and Justin went, and then subtracting their distances.

Cheetah are ferrocious animals with very high top speeds. The figure below shows the speed of a cheetah with time. You can assume the motion is towards the positive \(x \) axis, then the ordinate in the plot will be \(v_x \) with abscissa \(t\) shown in Figure 2.9.28.

(a) Estimate how far cheetah goes in first \(30 \) seconds.

(b) Estimate acceleration of cheetah at the following instants: (i) \(t = 0\text{,}\) (ii) \(t = 5\text{ sec}\text{,}\) (iii) \(t = 17.5\text{ sec}\text{,}\) and (iv) \(t = 30\text{ sec}\text{.}\) At each instant, indicate both the magnitude and direction of the acceleration.

Figure 2.9.28. Figure for Problem 2.9.27.

If you are curious about Cheetahs, see this Houston Zoo Video on the Youtube.

Hint 1 (a)

(a) You can approximate the area of interest as made up areas of triangles and rectangles.

Hint 2 (b)

(b) You can estimate instantaneous acceleration from the slope of the tangents.


(a) \(975\text{ m}\text{,}\) (b) (i)\(13.33\text{ m/s}^2\text{,}\) (ii) \(2.82\text{ m/s}^2\text{,}\) (iii) \(0 \text{,}\) (iv) \(-2.44\text{ m/s}^2\text{.}\)

Solution 1 (a)

(a) Our answers will differ since we may break up the area between \(t=0\) and \(t=30\text{ sec}\) in different ways. My area will come from the areas shaded in Figure 2.9.29.

Figure 2.9.29. Figure for part (a) of Problem 2.9.27.

The areas sum up to

\begin{align*} \text{triangles } \amp +\text{ rectangles} = \\ \amp (75+50+50) + (750+50) = 975\text{ m}. \end{align*}
Solution 2 (b)

(b) Our answers will differ since we may draw tangents a little differently. Figure 2.9.30 contains the tangents at the four instants.

Figure 2.9.30. Figure for part (b) of Problem 2.9.27.

The slopes of these tangents give the acceleration at the corresponding instants.

\begin{align*} \amp \text{(i)}\ \ a_x(\text{at }0) = \dfrac{40-0}{3-0} = 13.33\text{ m/s}^2, \\ \amp \text{(ii)}\ \ a_x(\text{at }5) = \dfrac{40-16}{8.5-0} = 2.82\text{ m/s}^2, \\ \amp \text{(iii)}\ \ a_x(\text{at }17.5) = 0, \\ \amp \text{(iv)}\ \ a_x(\text{at }30) = \dfrac{18-40}{35-26} = -2.44\text{ m/s}^2. \end{align*}

The value at \(t= 30\text{ sec}\) is negative because the cheetah, while moving towards positive \(x \) axis, is slowing down.

Figure 2.9.32 shows speed of a train between two stations. Assume a straight track between the two stations.

(a) What is the distance between the two stations?

(b) Supposing \(x \) axis is pointed from A towards B, plot the acceleration \(a_x \) versus \(t \text{.}\)

(c) Using your acceleraiton values, find the distance travelled in the three durations (i) from \(0 \) to \(t = 2\text{ sec} \text{,}\) (ii) from \(t = 2\text{ sec} \)to \(t = 5\text{ sec} \text{,}\) and (iii) from \(t = 5\text{ sec} \)to \(t = 6\text{ sec} \text{.}\) Verify that your answer to this part agrees with your answer to part (a)

Figure 2.9.32. Figure for Problem 2.9.31.

(a) Area under the curve, (b) Slopes, (c) Use constant acceleration in the three segments.


(a) \(10,800\text{ m}\text{,}\) (b) three acceleration values \(\dfrac{1}{3}\text{ m/s}^2\text{,}\) \(0\text{,}\) \(-\dfrac{2}{3}\text{ m/s}^2\text{,}\) (c) \(2400\text{ m} + 7200\text{ m} + 1200\text{ m} = 10,800\text{ m}\text{.}\)

Solution 1 (a)

(a) Since the motion does not have a turn around in direciton, the speed plot is same as velocity plot. Therefore, we can get the distance traveled by area-under-the-curve. We also convert the time into seconds.

\begin{equation*} d = \dfrac{1}{2}\times 120\times 40 + 180\times 40 + \dfrac{1}{2}\times 60\times 40 = 10,800\text{ m}. \end{equation*}
Solution 2 (b)

(b) From the slopes we find three values of accelerations.

\begin{align*} \amp a_{i,x} = \dfrac{40}{120} = \dfrac{1}{3}\text{ m/s}^2, \\ \amp a_{ii,x} = 0, \\ \amp a_{iii,x} = \dfrac{-40}{60} = -\dfrac{2}{3}\text{ m/s}^2. \end{align*}

We plot these in Figure 2.9.33.

Figure 2.9.33. Figure for part (b) of Problem 2.9.31.
Solution 3 (c)

(c) (i) With \(a_{x} = \dfrac{1}{3}\text{ m/s}^2\text{,}\) \(v_{i,x}=0\text{,}\) and \(\Delta t = 120\text{ sec}\text{,}\) we get

\begin{equation*} \Delta x_1 = 0 + \dfrac{1}{2}\times \dfrac{1}{3} \times 120^2 = 2400\text{ m}. \end{equation*}

(ii) This is a constant velocity motion.

\begin{equation*} \Delta x_2 = v_2 \Delta t = 40\times 180 = 7200\text{ m}. \end{equation*}

(iii) Finally, in this part, we have \(a_{x} = -\dfrac{2}{3}\text{ m/s}^2\text{,}\) \(v_{i,x}=40\text{ m/s}\text{,}\) and \(\Delta t = 60\text{ sec}\text{.}\) These give us

\begin{equation*} \Delta x_3 = 40\times 60 - \dfrac{1}{2}\times \dfrac{2}{3} \times 60^2 = 1200\text{ m}. \end{equation*}

Therefore, the total

\begin{equation*} \Delta x = 2400 + 7200 + 1200 = 10,800\text{ m}. \end{equation*}

Two stones are dropped from the roof of a \(50\text{-m}\) tall building. The first stone is released smoothly from rest, and after \(0.5\text{ sec}\) later, the second stone is thrown vertically down with an intial speed such that they reach the ground at the same time.

(a) With what initial speed was the second stone thrown?

(b) With what speed did the first stone hit the ground?

(c) With what speed did the second stone hit the ground?


Both stones follown free fall motion related in time by \(t_1 = t_2 + 0.5 \text{.}\) Set up two sets of equations, one for each stone and solve them simultaneously.


(a) \(5.29\text{ m/s} \text{,}\) (b) \(31.4\text{ m/s}\text{,}\) (c) \(31.78\text{ m/s} \text{.}\)

Solution (a),(b),(c)

(a), (b) , (c), all three parts can be solved by setting up two sets of equations, one set for each stone, and relation \(t_1 = t_2 + 0.5 \) for the times of the two stones. Let us use positive \(y\) axis up to keep track of signs with origin at the place the motions started. That will make \(y_i = 0\) and \(y_f = -50 \text{ m}\text{.}\) Let us use subscript 1 for stone 1 and 2 for stone 2, and drop \(y\) from subscripts.

For stone 1, we have the following two equations.

\begin{align} -50 \amp = 0 - \dfrac{1}{2}\, 9.81\, t_1^2,\label{eq-Two-Stones-Dropped-at-Different-Times-1a}\tag{2.9.1}\\ v_{1f} \amp= 0 - 9.81\, t_1,\label{eq-Two-Stones-Dropped-at-Different-Times-1b}\tag{2.9.2} \end{align}

For stone 2, we have the following two equations.

\begin{align} -50 \amp= v_{2i}\, t_2 - \dfrac{1}{2}\, 9.81\, t_2^2,\label{eq-Two-Stones-Dropped-at-Different-Times-2a}\tag{2.9.3}\\ v_{2f} \amp = v_{2i} - 9.81\, t_2,\label{eq-Two-Stones-Dropped-at-Different-Times-2b}\tag{2.9.4} \end{align}


\begin{equation*} t_2 = t_1 - 0.5. \end{equation*}

From Eq. (2.9.1), we solve for \(t_1\) and keep the positive root as our answer.

\begin{equation*} t_1 = 3.2\text{ sec}. \end{equation*}


\begin{equation*} t_2 = 2.7\text{ sec}. \end{equation*}

Using \(t_1\) in Eq. (2.9.2) give,

\begin{equation*} v_{1f} = - 9.81 \times 3.2 = -31.4\text{ m/s}. \end{equation*}

Using \(t_2\) in Eq. (2.9.3) gives

\begin{equation*} v_{2i} = \dfrac{-50 + 4.9 \times 2.7^2 } { 2.7 } = -5.29\text{ m/s}. \end{equation*}

Using \(t_2\) and \(v_{2i}\) in Eq. (2.9.4) gives

\begin{equation*} v_{2f} = -5.29 - 9.81 \times 2.7 = -31.78\text{ m/s}. \end{equation*}

Two balls are launched straight up at different times from a tennis ball launching machine. The second ball is launched at the time the first ball is at the top of its flight. If the launching speed of the balls are \(20\text{ m/s}\text{,}\) (a) where would the two balls hit each other and (b) what will be the time when they hit the ball if \(t=0\) at the instant of the launch of the second ball? Assume the effect of the air resistance to be negligible on the motion.


First find the height the first ball will fly before falling back.


(a) \(15.3\text{ m}\text{,}\) (b) \(1.02\text{ s}\text{.}\)

Solution (a),(b)

Let \(y=0\) at the launch point with positive \(y\) axis pointed up. The first ball reaches \(y=h\) before falling with

\begin{equation*} y = h = \dfrac{-v_i^2}{2g} = \dfrac{-20^2}{2\times 9.81} = 20.39\text{ m}. \end{equation*}

Now, ball 1 moves in the negative \(y\) direction at \(t=0\) starting at \(y_{1i}=20.39\text{ m}\) with \(v_{1i}=0\) and ball 2 moves in the positive \(y\) direction \(y_{2i}=0\) starting with \(v_{2i}=20\text{ m/s}\text{.}\) We want \(y\) where they hit each other, i.e., \(y_{1f} = y_{2f} = y\text{.}\) Let \(t\) be the duration since ball 2 was launched. For the two equations we get

\begin{gather*} \text{ ball 1: } y - 20.39 = - \dfrac{1}{2}\times 9.81\, t^2\\ \text{ ball 2: } y = 20\, t - \dfrac{1}{2}\times 9.81\, t^2 \end{gather*}

Subtracting first equation from the second we get

\begin{equation*} 20.39 = 20\, t\ \ \Longrightarrow\ \ t = 1.02\text{ s}. \end{equation*}

Using this \(t\) we get

\begin{equation*} y = 20.39 - \dfrac{1}{2}\times 9.81\times 1.02^2 = 15.3\text{ m}. \end{equation*}

A hockey puck is shot at \(40\text{ m/s}\text{.}\) Initially, the puck decelerates at \(3\text{ m/s}^2\text{,}\) but after \(50\text{ m}\text{,}\) the puck runs into a rough patch, where the puck has constant deceleration \(12\text{ m/s}^2\) and comes to rest.

(a) How far from the initial place does the puck come to rest?

(b) How much time did it take for the puck to come to rest?


You will need to set up two sets of equations with final velocity of segment 1 is equal to the initial velocity of segment 2.


(a) \(104.3\text{ m}\text{,}\) (b) \(4.31\text{ sec} \)

Solution (a),(b)

(a) and (b):

There are two constant-acceleration segments in this problem, such that final velocity and final position of the first segment is the initial velocity and initial position of the second segment. Let us denote the quantities of the first and second segments by subscripts 1 and 2, respectively.

We point positive \(x \) axis from the initial point of the first segment towards the final point where the puck comes to rest. Using this coordinate system, we have the following knowns and unknowns for segment 1.

First Segment:

Suppressing units in calculations, we list knowns and unknowns.

\begin{align*} \amp a_1 = - 3,\ \ t_1 = ?, \\ \amp x_{1,i} = 0,\ \ v_{1,i} = 40,\ \ x_{1,f} = 50,\ \ v_{1,f} = ?, \end{align*}

The equation \(v_f^2 = v_i^2 + 2 a x \) gives

\begin{equation*} v_{1,f}^2 = 40^2 + 2 \times (-3) \times 50 = 1600 - 300 = 1300. \end{equation*}

This gives

\begin{equation*} v_{1,f} = \pm 36.1\text{ m/s}. \end{equation*}

Since the puck is moving in the same direction as the direction of the positive \(x \) axis, the positive root is the right choice here.

\begin{equation*} v_{1,f} = + 36.1\text{ m/s}. \end{equation*}

Now, using \(v_f = v_i + a t\text{,}\) we get

\begin{equation*} t = \dfrac{v_f - v_i}{a} = \dfrac{36.1 - 40}{-3} = 1.3\text{ sec}. \end{equation*}

Second Segment:

Let us use \(t_2 \) for \(t_{2,f} - t_{2,i}\) and place another origin at \(x_{2,i}\text{.}\)

\begin{align*} \amp a_2 = - 12,\ \ t_2 = ?, \\ \amp x_{2,i} = 0,\ \ v_{2,i} = 36.1,\ \ x_{2,f} = ?,\ \ v_{2,f} = 0, \end{align*}

Using \(v_f = v_i + a t\text{,}\) we get \(t\)

\begin{equation*} t_2 = \dfrac{v_{2,f} - v_{2,i}}{a_2} = \dfrac{0 - 36.1}{-12} = 3.01\text{ sec}. \end{equation*}

Using \(x = v_i t +\dfrac{1}{2}at^2\text{,}\) we get

\begin{equation*} x_{2,f} = 36.1\times 3.01 +\dfrac{1}{2}\times(-12) \times 3.01^2 = 54.3\text{ m}. \end{equation*}

(a) The total distance is \(50 + 54.3 = 104.3\text{ m}\text{.}\)

(b) The total time is

\begin{equation*} t_1 + t_2 = 1.3 + 3.01 = 4.31\text{ sec}. \end{equation*}

A parachuter drops off an air plane with zero vertical speed. Suppose magnitude of acceleration of the parachuter varies linearly in time, from \(g \) at \(t=0\) to zero at \(t=\tau \text{.}\)

(a) What is the speed at an arbitrary instant \(t \) before the acceleration is zero?

(b) What is the speed at the instant \(\tau \text{?}\)

(c) How far has he fallen in time \(\tau \text{?}\)


(a) Integrate \(a_y\text{.}\) (b) Evaluate at \(t = \tau\) . (c) Integrate \(v_y\text{.}\)


(a) \(\dfrac{g}{2\tau}\, t^2 - g t \text{,}\) (b) \(\dfrac{1}{2} g\tau\text{,}\) (c) \(\dfrac{1}{3}\,g \tau^2\text{.}\)

Solution 1 (a)

(a) We would find velocity by integrating the acceleration, and then get the speed from the absolute value. The magnitude of acceleration drops from \(g \) to zero in time \(\tau\text{.}\) Therefore, we have the following formula for the magnitude of the acceleration at instant \(t\text{.}\)

\begin{equation*} a(t) = \begin{cases} \dfrac{g}{\tau}\, |\tau - t| \amp t \le \tau\\ 0 \amp t \gt \tau \end{cases} \end{equation*}

Let positive \(y \) axis be pointed up so that

\begin{equation*} a_y(t) = \begin{cases} \frac{g}{\tau} t - g \amp t \le \tau\\ 0 \amp t \gt \tau \end{cases} \end{equation*}

Now, we integrate this from \(t = 0\) to \(t=t\text{.}\) With \(v_y=0\) at \(t = 0\text{.}\)

\begin{equation} v_y(t) - 0 = \int_0^{t} a_y(t)\, dt = -\dfrac{g}{2\tau}\, t^2 - g t.\label{eq-Varying-Acceleration-During-Parachute-Jump-vy}\tag{2.9.5} \end{equation}

Therefore, speed at arbitrary instant is

\begin{equation*} v = |\dfrac{g}{2\tau}\, t^2 - g t|. \end{equation*}
Solution 2 (b)

(b) Speed at instant \(\tau\) will be obtained by setting \(t= \tau\) in the answer for part (a) and taking the absolute value.

\begin{equation*} v = \dfrac{1}{2} g\tau. \end{equation*}
Solution 3 (c)

(c) Integrating \(v_y(t) \) will give \(\Delta y\text{.}\)

\begin{equation*} y(t) - 0 = \int_0^{\tau} v_y(t)\, dt = -\dfrac{g}{6\tau}\, \tau^3 -\dfrac{1}{2}g \tau^2 = -\dfrac{1}{3}\,g \tau^2. \end{equation*}

The magnitude of this gives the distance fallen.

\begin{equation*} h = |y(t)| = \dfrac{1}{3}\,g \tau^2. \end{equation*}

Consider an object falling vertically. In a coordinate system with the positive \(y \) axis pointed down, the acceleration \(a_y\) has the following expression. Let \(y=0\) at \(t=0\text{.}\)

\begin{equation*} a_y = g - \gamma v_y, \end{equation*}

where \(g \) is acceleration due to gravity, and \(\gamma \) is a constant, called the damping constant.

(a) Show that at the instant the acceleration is zero, the speed is

\begin{equation*} v = \dfrac{g}{\gamma}. \end{equation*}

The speed is called the terminal speed. We will denote it by \(v_T \text{.}\)

(b) Prove that velocity at arbitrary instant \(t \) is given by

\begin{equation*} v_y = v_T\left( 1 - e^{-\gamma\, t} \right). \end{equation*}

(c) Prove that the \(y \) coordinate at arbitrary instant \(t \) is given by

\begin{equation*} y = v_T\, t - \dfrac{v_T}{\gamma}\left( 1 - e^{-\gamma\, t} \right). \end{equation*}

(a) Set \(a_y=0\text{.}\) (b) Write \(a_y = dv_y/dt\text{,}\) then change variable to \(k = -\gamma v_y + g \) and integrate. (c) Integrate \(v_y\text{.}\)


Already given in problem statement.

Solution 1 (a)

(a) Setting \(a_y=0\) and \(v_y = v_T \text{,}\) we imediately get the answer.

\begin{equation*} v_T = \dfrac{g}{\gamma}. \end{equation*}
Solution 2 (b)

(b) We express \(a_y = dv_y/dt\) and rearrange to obtain

\begin{equation*} \dfrac{dv_y}{g - \gamma v_y} = dt. \end{equation*}

Integrating this from \(t=0\) to \(t =t\) with \(v_y(0)=0\) and \(v_y(t)=v_y\text{,}\) we get

\begin{equation*} \ln\left( \dfrac{| g - \gamma v_y|}{g} \right) = -\gamma t. \end{equation*}

Since \(g\) is larger than \(\gamma v_y \text{,}\) exponentiating and then rearranging gives.

\begin{equation*} v_y = \dfrac{g}{\gamma} \left( 1 - e^{-\gamma t}\right), \end{equation*}

which is

\begin{equation*} v_y = v_T\left( 1 - e^{-\gamma\, t} \right). \end{equation*}
Solution 3 (c)

(c) Integrating \(v_y\) gives the desired answer.

\begin{equation*} \int dy = \int v_y dt. \end{equation*}

Use \(v_y \) from (b) and carry out the integration.

\begin{equation*} y = v_T\, t - \dfrac{v_T}{\gamma}\left( 1 - e^{-\gamma\, t} \right). \end{equation*}

A runner runs on a straight track. Starting from rest, the runner runs for \(10\text{ sec}\) at a constant acceleration of \(0.4\text{ m/s}^2\text{,}\) then he decreases his acceleration to \(0.2\text{ m/s}^2\) for another \(12\text{ sec}\text{.}\) After that, he coasts for \(38\text{ sec}\) at the final speed. How far has he gone from the starting place?


No solution provided.

A car's acceleration changes with time according to

\begin{equation*} a(t) = 5.0\left( 1 - e^{-0.2\, t} \right). \end{equation*}

Here unit of acceleration is \(\text{m/s}^2\text{.}\) Starting from rest at \(t=0\text{,}\) how far will the car go in \(5.0\text{ sec}\text{?}\)


No solution provided.

A ball is tossed straight up with an unknown speed and round trip times (ACA and BCB) at two different heights marked A and B in the figure are observed.

With height \(H = 2\text{ m}\) between A and B, return times observed are \((ACA) = 2.8\text{ sec}\) and \((BCB) = 2.0\text{ sec}\text{.}\) From this data, find the value of \(g\text{.}\) (Don't assume \(g=9.81\text{ m/s}^2\text{.}\))


No solution provided.