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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 41

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be y0=0.

Solution:

The given known quantities are:a=−9.8 m/s2a=-9.8\:\text{m/s}^2; yo=0 my_o=0\:\text{m}; and voy=+15 m/sv_{oy}=+15\:\text{m/s}.

To compute for the displacement, we use the formula

Δy=voyt+12at2\Delta y=v_{oy}t+\frac{1}{2}at^2

and to compute for the final velocity, we use the formula

vfy=voy+atv_{fy}=v_{oy}+at

Part A

The displacement at t=0.500 st=0.500 \ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(0.500 s)+12(−9.8 m/s2)(0.500 s)2Δy=6.28 m  (Answer)\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(0.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right)^2 \\ \Delta y & =6.28\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=0.500 st=0.500 \ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(0.500 s)vfy=10.1 m/s  (Answer)\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right) \\ v_{fy} & =10.1\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The displacement at t=1.000 st=1.000 \ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(1.000 s)+12(−9.8 m/s2)(1.000 s)2Δy=10.1 m  (Answer)\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(1.000\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.000\:\text{s}\right)^2 \\ \Delta y & =10.1\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=1.000 st=1.000\ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(1.000 s)vfy=5.20 m/s  (Answer)\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.000\:\text{s}\right) \\ v_{fy} & =5.20\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The displacement at t=1.500 st=1.500\ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(1.500 s)+12(−9.8 m/s2)(1.500 s)2Δy=11.5 m  (Answer)\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(1.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.500\:\text{s}\right)^2 \\ \Delta y & =11.5\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=1.500 st=1.500\ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(1.500 s)vfy=0.300 m/s  (Answer)\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.500\:\text{s}\right) \\ v_{fy} & =0.300\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

The displacement at t=2.000 st=2.000\ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(2.000 s)+12(−9.8 m/s2)(2.000 s)2Δy=10.4 m  (Answer)\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(2.000\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.000\:\text{s}\right)^2 \\ \Delta y & =10.4\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=2.000 st=2.000\ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(2.000 s)vfy=−4.600 m/s  (Answer)\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.000\:\text{s}\right) \\ v_{fy} & =-4.600\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 38

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s2 for 7.00 s.

(a) What is his final velocity?

(b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save?

(c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

Solution:

We are given the following: v0=11.5 m/sv_0=11.5 \ \text{m/s} ; a=0.500 m/s2 a=0.500 \ \text{m/s}^2; and Δt=7.00 s \Delta t=7.00 \ \text{s}.

Part A

To solve for the final velocity, we are going to use the formula

vf=v0+atv_f=v_0+at

Substituting the given values:

vf=v0+atvf=11.5 m/s+(0.500 m/s2)(7.00 s)vf=15.0 m/s  (Answer)\begin{align*} v_f &=v_0+at\\ v_f&=11.5\ \text{m/s}+\left( 0.500\ \text{m/s}^2 \right)\left( 7.00\ \text{s} \right)\\ v_f&=15.0\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Let tconst t_{const} be the time it takes to reach the finish line without accelerating:

tconst=xv0tconst=300 m11.5 m/stconst=26.1 m/s\begin{align*} t_{const}&=\frac{x}{v_0}\\ t_{const}&=\frac{300\ \text{m}}{11.5\ \text{m/s}}\\ t_{const}&=26.1\ \text{m/s} \end{align*}

Now let dd be the distance traveled during the 7 seconds of acceleration. We know t=7.00 st=7.00 \ \text{s} so

d=v0t+12at2d=(11.5 m/s)(7.00 s)+12(0.500 m/s2)(7.00 s)2d=92.8 m\begin{align*} d&=v_0t+\frac{1}{2}at^2\\ d&=\left( 11.5\ \text{m/s} \right)\left( 7.00\ \text{s} \right)+\frac{1}{2}\left( 0.500\ \text{m/s} ^2\right)\left( 7.00\ \text{s} \right)^2\\ d&=92.8\ \text{m} \end{align*}

Let t′t' be the time it will take the rider at the constant final velocity to complete the race:

t′=x−dvt′=300 m−92.8 m15.0 m/st′=13.8 s\begin{align*} t'&=\frac{x-d}{v}\\ t'&=\frac{300\ \text{m}-92.8\ \text{m}}{15.0\ \text{m/s}}\\ t'&=13.8\ \text{s} \end{align*}

So the total time TT it will take the accelerating rider to reach the finish line is 

T=t+t′T=7.00 s+13.8 sT=20.8 s\begin{align*} T&=t+t'\\ T&=7.00\ \text{s}+13.8\ \text{s}\\ T&=20.8\ \text{s} \end{align*}

Finally, let T∗T^{*} be the time saved. So 

T∗=26.1 s−20.8 sT∗=5.3 s  (Answer)\begin{align*} T^{*}&=26.1\ \text{s}-20.8\ \text{s}\\ T^{*}&={\color{DarkGreen} 5.3\ \text{s}} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

For rider 2, we are given the following values: Δx′=295 m\Delta x'=295 \ \text{m} ; v′=11.8 m/sv'=11.8 \ \text{m/s}

Let t2 t_2 be the time it takes for rider 2 to reach the finish line.

We are going to use the formula

t2=Δx′v′t_2=\frac{\Delta x'}{v'}

Substituting the given values:

t2=x′v′t2=295 m11.8 m/st2=25.0 s\begin{align*} t_2 & =\frac{x'}{v'} \\ t_2 & =\frac{295\:\text{m}}{11.8\:\text{m/s}} \\ t_2 & =25.0\:\text{s} \end{align*}

The time difference is

time difference=t2−Ttime difference=25.0 s−20.817 stime difference=4.2 s  (Answer)\begin{align*} \text{time difference} & =t_2-T \\ \text{time difference} & =25.0\:\text{s}-20.817\:\text{s} \\ \text{time difference} & =4.2\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, he finishes 4.2 s after the winner.

When the other racer reaches the finish line, he has been traveling at 11.8 m/s for 4.2 seconds, so the other racer finishes

Δx=(11.8 m/s)(4.2 s)Δx=49.56 m  (Answer)\begin{align*} \Delta x & =\left(11.8\:\text{m/s}\right)\left(4.2\:\text{s}\right) \\ \Delta x & =49.56\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

behind the other racer.

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College Physics by Openstax Chapter 2 Problem 37

Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11.

(a) Calculate the average acceleration for such a dragster.

(b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time.

(c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.

Solution:

We are given the following: v0=0 m/sv_0=0\ \text{m/s} ; vf=145 m/sv_f=145 \ \text{m/s}; and Δt=4.45 sec\Delta t=4.45 \ \text{sec} .

Part A

To compute for the average acceleration aa, we are going to use the formula

a=ΔvΔt=vf−v0Δta=\frac{\Delta v}{\Delta t}=\frac{v_f-v_0}{\Delta t}

Substituting the given values, we have

a=vf−v0Δta=145 m/s−0 m/s4.45 sa=32.6 m/s2  (Answer)\begin{align*} a & =\frac{v_f-v_0}{\Delta t} \\ a & =\frac{145\:\text{m/s}-0\:\text{m/s}}{4.45\:\text{s}} \\ a & =32.6\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are given the following: a=32.6 m/s2a=32.6 \ \text{m/s}^2 ; v0=0 m/s v_0=0 \ \text{m/s}; and Δx=402 m\Delta x=402 \ \text{m} .

Since we do not have any information on time, we are going to use the formula

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

To compute for the final velocity, we have

vf=(v0)2+2aΔ xv_f=\sqrt{\left(v_0\right)^2+2a\Delta \:x}

Substituting the given values:

vf=(v0)2+2aΔ xvf=(0 m/s)2+2(32.6 m/s2)(402 m)vf=162 m/s  (Answer)\begin{align*} v_f & =\sqrt{\left(v_0\right)^2+2a\Delta \:x} \\ v_f & =\sqrt{\left(0\:\text{m/s}\right)^2+2\left(32.6\:\text{m/s}^2\right)\left(402\:\text{m}\right)} \\ v_f & =162\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The final velocity is greater than that used to find the average acceleration because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at the beginning, so it would not be accelerating at 32.6 m/s2 during the last few meters, but substantially less, and the final velocity would be less than  162 m/s162\:\text{m/s}.

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College Physics by Openstax Chapter 2 Problem 34

In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few life-threatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot’s speed upon impact was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m.

Solution:

We are given the following: x=3 mx=3\ \text{m}; v0=54 m/sv_0=54\ \text{m/s}; and vf=0 m/sv_f=0 \ \text{m/s}.

We are required to solve for the acceleration, and we are going to use the formula

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

Solving for acceleration aa in terms of the other variables:

a=(vf)2−(v0)22Δxa=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values:

a=(vf)2−(v0)22Δxa=(0 m/s)2−(54 m/s)22(3.0 m)a=−486 m/s2  (Answer)\begin{align*} a & = \frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0\:\text{m/s}\right)^2-\left(54\:\text{m/s}\right)^2}{2\left(3.0\:\text{m}\right)} \\ a & =-486\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The negative acceleration means that the pilot was decelerating at a rate of 486 m/s every second.

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