Tag Archives: acceleration vs time graph

College Physics by Openstax Chapter 2 Problem 64

(a) Take the slope of the curve in Figure 2.64 to find the jogger’s velocity at t=2.5 s. (b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 2.65.

Figure 2.64

Solution:

Part A

To find the slope at t=2.5 s, we need the position values at t= 0 s and t=5 s. When t=0 st = 0 \ \text{s}, x=0 mx = 0 \ \text{m}, and when t=5 st = 5 \ \text{s}, x=17.5 mx = 17.5 \ \text{m}. The velocity at t=2.5 s is

velocity=slopev=ΔxΔtv=x2−x1t2−t1v=17.5 m−0 m5 s−0 sv=3.5 m/s  (Answer)\begin{align*} \text{velocity} & =\text{slope} \\ \text{v} & =\frac{\Delta x}{\Delta t} \\ \text{v} & = \frac{x_2-x_1}{t_2-t_1} \\ \text{v} & = \frac{17.5\ \text{m}-0\ \text{m}}{5 \ \text{s}-0\ \text{s}} \\ \text{v} & =3.5 \ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\ \end{align*}

Part B

When t=10 st = 10 \ \text{s}, x=2.5 mx=2.5 \ \text{m}. Considering the points at t=5 s and t=10 s, the slope at 7.5 s is

velocity=slopev=ΔxΔtv=x2−x1t2−t1v=2.5 m−17.5 m10 s−5 sv=−3.0 m/s  (Answer)\begin{align*} \text{velocity} & =\text{slope} \\ \text{v} & =\frac{\Delta x}{\Delta t} \\ \text{v} & = \frac{x_2-x_1}{t_2-t_1} \\ \text{v} & = \frac{2.5\ \text{m}-17.5\ \text{m}}{10 \ \text{s}-5\ \text{s}} \\ \text{v} & =-3.0 \ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\ \end{align*}
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College Physics by Openstax Chapter 2 Problem 63

Construct the position graph for the subway shuttle train as shown in Figure 2.18(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

Figure 2.18

Solution:

The figure with the corresponding examples are shown on this page: https://openstax.org/books/college-physics/pages/2-4-acceleration#import-auto-id2590556

The position-vs-time graph of the train’s motion is also graphed in the first figure here: https://openstax.org/apps/archive/20210713.205645/resources/a697c43432cdf2d09a02df47d2b746283b841fcd

(a) Position of the train over time. Notice that the train’s position changes slowly at the beginning of the journey, then more and more quickly as it picks up speed. Its position then changes more slowly as it slows down at the end of the journey. In the middle of the journey, while the velocity remains constant, the position changes at a constant rate. (b) The velocity of the train over time. The train’s velocity increases as it accelerates at the beginning of the journey. It remains the same in the middle of the journey (where there is no acceleration). It decreases as the train decelerates at the end of the journey. (c) The acceleration of the train over time. The train has positive acceleration as it speeds up at the beginning of the journey. It has no acceleration as it travels at constant velocity in the middle of the journey. Its acceleration is negative as it slows down at the end of the journey.

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College Physics by Openstax Chapter 2 Problem 65

A graph of v(t) is shown for a world-class track sprinter in a 100-m race. (See Figure 2.67). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at t=5 s? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

A graph of v(t) is shown for a world-class track sprinter in a 100-m race.
Figure 2.67

Solution:

Part A

To find the average velocity over the straight line graph of the velocity vs time shown, we just need to locate the midpoint of the line. In this case, the average speed for the first 4 seconds is

vave=6 m/s  (Answer)v_{\text{ave}}=6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

Looking at the graph, the velocity at exactly 5 seconds is 12 m/s.

Part C

If we are given the velocity-time graph, we can solve for the acceleration by solving for the slope of the line.

Consider the line from time zero to time, t=4 seconds. The slope, or acceleration, is

a=slope=12 m/s−0 m/s4 s=3 m/s2  (Answer)a=\text{slope}=\frac{12\:\text{m/s}-0\:\text{m/s}}{4\:\text{s}}=3\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part D

For the first 4 seconds, the distance traveled is equal to the area under the curve.

distance=12(4 sec⁡)(12 m/s)=24 m\text{distance}=\frac{1}{2}\left(4\:\sec \right)\left(12\:\text{m/s}\right)=24\:\text{m}

So, the sprinter traveled a total of 24 meters in the first 4 seconds. He still needs to travel a distance of 76 meters to cover the total racing distance. At the constant rate of 12 m/s, he can run the remaining distance by

t=distancevelocity=76 m12 m/s=6.3 sec⁡\text{t}=\frac{\text{distance}}{\text{velocity}}=\frac{76\:\text{m}}{12\:\text{m/s}}=6.3\:\sec

Therefore, the total time of the sprint is

ttotal=4 sec⁡+6.3 sec⁡=10.3 sec⁡  (Answer)\text{t}_{\text{total}}=4\:\sec +6.3\:\sec =10.3\:\sec \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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College Physics by Openstax Chapter 2 Problem 36

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of  0.150 m/s2 as it goes through.  The station is 210 m long.

(a) How long is the nose of the train in the station?

(b) How fast is it going when the nose leaves the station?

(c) If the train is 130 m long, when does the end of the train leave the station?

(d) What is the velocity of the end of the train as it leaves?

Solution:

Part A

We are given the following: v0=22.0 m/sv_0=22.0\:\text{m/s}; a=−0.150 m/s2a=-0.150\:\text{m/s}^2; and Δx=210 m\Delta x=210\:\text{m}

We are required to solve for time, tt. We are going to use the formula

Δx=v0t+12at2\Delta x=v_0t+\frac{1}{2}at^2

Substituting the given values, we have

Δx=v0t+12at2210 m=(22.0 m/s)t+12(−0.150 m/s2)t2\begin{align*} \Delta x & =v_0t+\frac{1}{2}at^2 \\ 210\:\text{m} & =\left(22.0\:\text{m/s}\right)t+\frac{1}{2}\left(-0.150\:\text{m/s}^2\right)t^2 \end{align*}

If we simplify and rearrange the terms into a general quadratic equation, we have

0.075t2−0.22t+210=00.075t^2-0.22t+210=0

Solve for tt using the quadratic formula. We are given a=0.075; b=−22t; c=210a=0.075;\:b=-22t;\:c=210.

t=−b±b2−4ac2at=−(−22)±(−22)2−4(0.075)(210)2(0.075)\begin{align*} t & =\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ t& =\frac{-\left(-22\right)\pm \sqrt{\left(-22\right)^2-4\left(0.075\right)\left(210\right)}}{2\left(0.075\right)}\\ \end{align*}

There are two values of tt that can satisfy the quadratic equation.

t=9.88 sandt=283.46 st=9.88\ \text{s} \qquad \text{and} \qquad t=283.46 \ \text{s}

Discard t=283.46 st=283.46 \ \text{s} as it can be seen from the problem that this is not a feasible solution. So, we have

t=9.88 s  (Answer)t=9.88\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

We have the same given values from part a. We are going to solve vfv_f in the formula

(vf)2=(v0)2+2aΔx\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

So we have

(vf)2=(v0)2+2aΔxvf=(v0)2+2aΔxvf=(22.0 m/s)2+2(−0.150 m/s2)(210 m)vf=20.6 m/s  (Answer)\begin{align*} \left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\ v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\ v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(210\:\text{m}\right)} \\ v_f & =20.6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

We are given the same values as in the previous parts, except that for the value of Δx\Delta x since we should incorporate the length of the train. For the distance, Δx\Delta x, we have

Δx=210 m+130 mΔx=340 m\begin{align*} \Delta x & =210\:\text{m}+130\:\text{m} \\ \Delta x & = 340 \ \text{m} \end{align*}

To solve for time tt, we are going to use the formula

Δx=vot+12at2\Delta x=v_ot+\frac{1}{2}at^2

If we rearrange the formula into a general quadratic equation and solve for tt using the quadratic formula, we come up with

t=−v0±v02+2axat=\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a}

Substituting the given values:

t=−v0±v02+2axat=−(22.0 m/s)±(22.0 m/s)2+2(−0.150 m/s2)(340 m)−0.150 m/s2t=16.4 s  (Answer)\begin{align*} t & =\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a} \\ t & =\frac{-\left(22.0\:\text{m/s}\right)\pm \sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)}}{-0.150\:\text{m/s}^2} \\ t &=16.4\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

We have the same given values. We are going to solve for vfv_f in the equation

(vf)2=(v0)2+2aΔx\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Substituting the given values:

(vf)2=(v0)2+2aΔxvf=(v0)2+2aΔxvf=(22.0 m/s)2+2(−0.150 m/s2)(340 m)vf=19.5 m/s  (Answer)\begin{align*} \left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\ v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\ v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)} \\ v_f & =19.5\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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