Tag Archives: acceleration

Problem 6-17: The acceleration due to gravity at the position of a satellite located above the Earth

What percentage of the acceleration at Earth’s surface is the acceleration due to gravity at the position of a satellite located 300 km above Earth?

Solution:

The acceleration due to gravity of a body and the Earth is given by the formula

g=GMr2g= G \frac{M}{r^2}

where GG is the gravitational constant, MM is the mass of the Earth, and rr is the distance of the object to the center of the Earth. We know that the approximate radius of the Earth is r=6.3781×106 mr=6.3781 \times 10^6 \ \text{m} .

The percentage of the acceleration at 300 km above the Earth of the acceleration due to gravity at Earth’s surface is

(GMr2)2(GMr2)1×100%\displaystyle \frac{\left( \frac{GM}{r^2} \right)_2}{\left( \frac{GM}{r^2} \right)_1} \times 100\%

Note that the subscript 2 indicates the satellite located 300 km above the Earth, and the subscript 1 indicates the object at the Earth’s surface. Also, from the expression above, we can cancel GG and MM from the numerator and denominator because these are constants. So, we are down to

(1r2)2(1r2)1×100%=(r2)1(r2)2×100%\frac{\left( \frac{1}{r^2} \right)_2}{\left( \frac{1}{r^2} \right)_1} \times 100\% = \frac{\left( r^2 \right)_1}{\left( r^2 \right)_2} \times 100\%

Substituting the values, we have

(r2)1(r2)2×100%=(6.3781×106 m)2(6.3781×106 m+300×103 m)2×100%=91.2172%=91.2%  (Answer)\begin{align*} \frac{\left( r^2 \right)_1}{\left( r^2 \right)_2} \times 100\% & = \frac{\left( 6.3781 \times 10^6 \ \text{m} \right)^{2}}{\left( 6.3781 \times 10^6 \ \text{m}+300 \times 10^{3} \ \text{m} \right)^{2}} \times 100\% \\ \\ & = 91.2172\% \\ \\ & = 91.2\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The percentage of the acceleration at the Earth’s surface of the acceleration due to gravity at the position of a satellite located 300 km above the Earth is about 91.2%.

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College Physics by Openstax Chapter 2 Problem 43

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?

Solution:

It is our assumption that the player attempts to get the ball at the top where the velocity is zero.

The given are the following: vfy=0 m/sv_{fy}=0 \ \text{m/s}; Δy=1.25 m\Delta y=1.25 \ \text{m}; and a=−9.80 m/s2a=-9.80 \ \text{m/s}^2.

We are required to solve for the initial velocity v0yv_{0y} of the player. We are going to use the formula

(vfy)2=(voy)2+2aΔy\left(v_{fy}\right)^2=\left(v_{oy}\right)^2+2a\Delta y

Solving for voyv_{oy} in terms of the other variables:

voy=(vfy)2−2aΔyv_{oy}=\sqrt{\left(v_{fy}\right)^2-2a\Delta y}

Substituting the given values:

voy=(vfy)2−2aΔyvoy=(0 m/s)2−2(−9.80 m/s2)(1.25 m)voy=4.95 m/s  (Answer)\begin{align*} v_{oy} & =\sqrt{\left(v_{fy}\right)^2-2a\Delta y} \\ v_{oy} & = \sqrt{\left(0\:\text{m/s}\right)^2-2\left(-9.80\:\text{m/s}^2\right)\left(1.25\:\text{m}\right)} \\ v_{oy} & =4.95 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 40

(a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration.

(b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?

Solution:

Part A

There are two parts to the race and must be treated separately since acceleration is not uniform over the race. We will divide the race into Δx1\Delta x_1 (while accelerating) and Δx2\Delta x_2 (with constant speed), where Δx1+Δx2=100 m\Delta x_1 + \Delta x_2 = 100 \ \text{m} .

For Δx1\Delta x_1:

During the accelerating period, we are going to use the formula Δx=v0t+12at2\Delta x=v_0t+\frac{1}{2}at^2, since we know that a=Δvt=vmax−v0t=vmaxt\displaystyle a=\frac{\Delta v}{t}=\frac{v_{max}-v_0}{t}=\frac{v_{max}}{t}; and t=3.00 st=3.00 \ \text{s}.

Δx=v0t+12at2Δx1=0+12at2Δx1=12at2Δx1=12(vmaxt)t2Δx1=12(vmax)tΔx1=12(vmax)(3.00 s)Δx1=1.5vmax\begin{align*} \Delta x & =v_0t+\frac{1}{2}at^2 \\ \Delta x_1 & =0+\frac{1}{2}at^2 \\ \Delta x_1 & =\frac{1}{2}at^2 \\ \Delta x_1 & =\frac{1}{2}\left(\frac{v_{max}}{t}\right)t^2 \\ \Delta x_1 & =\frac{1}{2}\left(v_{max}\right)t \\ \Delta x _1&=\frac{1}{2}\left(v_{max}\right)\left(3.00\:\text{s}\right) \\ \Delta x _1&=1.5v_{max} \end{align*}

When the speed is constant, t=6.69 st=6.69 \ \text{s}, so

Δx2=vmaxtΔx2=vmax(6.69 s)Δx2=6.69vmax\begin{align*} \Delta x_2 & = v_{max}t \\ \Delta x_2 & = v_{max}\left(6.69\:\text{s}\right) \\ \Delta x_2 & =6.69v_{max} \end{align*}

Plugging-in the two equations in the equation Δx1+Δx2=100 m\Delta x_1 + \Delta x_2 = 100 \ \text{m} .

Δx1+Δx2=100 m1.5vmax+6.69vmax=100 m8.19 vmax=100vmax=1008.19vmax=12.2 m/s  (Answer)\begin{align*} \Delta x_1 + \Delta x_2 & = 100 \ \text{m} \\ 1.5v_{max} + 6.69v_{max} & =100 \ \text{m} \\ 8.19\:v_{max} & =100 \\ v_{max} & =\frac{100}{8.19} \\ v_{max} & =12.2\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, his acceleration can be computed using the formula

a=vmaxta=\frac{v_{max}}{t}

Plugging in the given values

a=vmaxta=12.2 m/s3.00 sa=4.07 m/s2  (Answer)\begin{align*} a & =\frac{v_{max}}{t} \\ a & = \frac{12.2\:\text{m/s}}{3.00\:\text{s}} \\ a & = 4.07\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Similar to part (a), we can plug in the different values for time and total distance:

Δx1+Δx2=2001.5 vmax+(19.30−3.00)vmax=2001.5 vmax+16.30vmax=20017.80vmax=200vmax=20017.80vmax=11.2 m/s  (Answer)\begin{align*} \Delta x_1+ \Delta x_2 & =200 \\ 1.5\:v_{max}+\left(19.30-3.00\right)v_{max} & =200 \\ 1.5\:v_{max}+16.30v_{max} & =200 \\ 17.80v_{max} & =200 \\ v_{max} & =\frac{200}{17.80} \\ v_{max} & = 11.2\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 39

In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, of 183.58 mi/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?

Solution:

There are two parts to the race: an acceleration part and a constant speed part.

For the acceleration part:

We are given the following values: v0=0 mphv_0=0 \ \text{mph} ; vf=60 mph v_f=60 \ \text{mph}; and Δt=4.00 s\Delta t=4.00 \ \text{s} .

First, we need to determine how long (both in distance and time) it takes the motorcycle to finish accelerating. During acceleration, the value of the acceleration is given by

a=60 mph4 sa=\frac{60\:\text{mph}}{4\:\text{s}}

To compute for the time it takes to reach its maximum velocity, we are going to use the formula

vf=v0+atv_f=v_0+at

Solving for time tt in terms of the other variables

t=vf−v0at=\frac{v_f-v_0}{a}

Substituting the given values to solve for t1t_1, the time it takes to accelerate from rest to maximum velocity:

t1=vf−v0at1=183 mph−0 mph(60 mph4 s)t1=12.2 s\begin{align*} t_1 & =\frac{v_f-v_0}{a} \\ t_1 & =\frac{183\:\text{mph}-0\:\text{mph}}{\left(\frac{60\:\text{mph}}{4\:\text{s}}\right)} \\ t_1 & =12.2\:\text{s} \end{align*}

Since we have a constant acceleration, the distance traveled Δx1\Delta x_1 during this period is computed using the formula

Δx1=vavetΔx1=(vf+v02)t\begin{align*} \Delta x_1 & =v_{ave}t \\ \Delta x_1 & =\left(\frac{v_f+v_0}{2}\right)t \\ \end{align*}

Substituting the given values:

Δx1=(vf+v02)tΔx1=(183 mph+0 mph2)(12.2 s)Δx1=(91.5 mph)(1 hr3600 s)(12.2 s)Δx1=0.31 mi\begin{align*} \Delta x_1 & =\left(\frac{v_f+v_0}{2}\right)t \\ \Delta x_1 & =\left(\frac{183\:\text{mph}+0\:\text{mph}}{2}\right)\left(12.2\:\text{s}\right) \\ \Delta x_1 & =\left(91.5\:\text{mph}\right)\left(\frac{1\:\text{hr}}{3600\:\text{s}}\right)\left(12.2\:\text{s}\right) \\ \Delta x_1 & =0.31\:\text{mi} \end{align*}

For the constant speed part:

For the next part of the motion, the speed is constant.

We are given the following values: Δx2=5.0 mi−0.3 mi=4.7 mi\Delta x_2=5.0\:\text{mi}-0.3\:\text{mi}=4.7\:\text{mi} .

We are going to solve t2t_2, the time spent on the course at max speed using the formula

Δx2=vmaxt2\Delta x_2=v_{max}t_2

Solving for t2t_2 in terms of the other variables:

t2=Δx2vmaxt_2=\frac{\Delta x_2}{v_{max}}

Substituting the given values:

t2=Δx2vmaxt2=4.7 mi183 mpht2=(0.026 h)(3600 s1 h)t2=92 s\begin{align*} t_2 & =\frac{\Delta x_2}{v_{max}} \\ t_2 & =\frac{4.7\:\text{mi}}{183\:\text{mph}} \\ t_2 & =\left(0.026\:\text{h}\right)\left(\frac{3600\:\text{s}}{1\:\text{h}}\right) \\ t_2 &=92\:\text{s} \end{align*}

For the whole course:

So, the total time is

ttotal=t1+t2ttotal=12.2 s+ 92 sttotal=104 s  (Answer)\begin{align*} t_{total}&=t_1+t_2 \\ t_{total}& =12.2\:\text{s}+\:92\:\text{s} \\ t_{total}& =104\:\text{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 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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College Physics by Openstax Chapter 2 Problem 35

Consider a grey squirrel falling out of a tree to the ground.

(a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m.

(b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.

Solution:

Part A

We are given the following: v0=0 m/sv_0=0 \ \text{m/s}; a=−9.80 m/s2a=-9.80 \ \text{m/s}^2; and Δx=−3.0 m\Delta x=-3.0\ \text{m}.

Note that the acceleration is due to gravity and its value is constant at a=−9.80 m/s2a=-9.80 \ \text{m/s}^2. Also, the distance xx, is negative because of the direction of motion. For free-fall, downward motion is considered negative. To solve for the velocity just before it hits the ground, we will solve vfv_fin the formula

(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=(0 m/s)2+2(−9.80 m/s2)(−3.0 m)vf=7.7 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(0\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(-3.0\:\text{m}\right)} \\ v_f & = 7.7\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are given the following: vf=0 m/sv_f=0 \ \text{m/s}; v0=7.7 m/sv_0=7.7 \ \text{m/s}; and Δx=0.02 m\Delta x=0.02 \ \text{m}

To solve for the acceleration we shall use the same formula as that in Part A. Solving for the acceleration aa:

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−(7.7 m/s)22(0.02 m)a=−1.5×103 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(7.7\:\text{m/s}\right)^2}{2\left(0.02\:\text{m}\right)} \\ a & =-1.5\times 10^3\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

This is approximately 3 times the deceleration of the airman from the previous problem, who was falling from thousands of meters high.

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College Physics by Openstax Chapter 2 Problem 33

An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m.

a) What is his deceleration?

b) How long does the collision last?

Solution:

We are given the following: v0=7.50 m/sv_0=7.50\:\text{m/s}; vf=0.00 m/sv_f=0.00\:\text{m/s}; andΔx=0.350 m\Delta x=0.350\:\text{m}.

Part A

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 the 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−(7.50 m/s)22(0.350 m)a=−80.4 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(7.50\:\text{m/s}\right)^2}{2\left(0.350\:\text{m}\right)} \\ a & =-80.4\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are going to use the formula

Δx=vavet\Delta x=v_{ave}t

Since vave=vf+v02v_{ave}=\frac{v_f+v_0}{2}, we can write the formula as

Δx=vf+v02⋅t\Delta x=\frac{v_f+v_0}{2}\cdot t

Solving for time tt in terms of the other variables:

 t=2Δxvf+v0\:t=\frac{2\Delta x}{v_f+v_0}

Substituting the given values:

t=2Δxvf+v0t=2(0.350 m)0 m/s+7.50 m/st=9.33×10−2 s  (Answer)\begin{align*} t&=\frac{2\Delta x}{v_f+v_0} \\ t&=\frac{2\left(0.350\:\text{m}\right)}{0\:\text{m/s}+7.50\:\text{m/s}} \\ t& =9.33\times 10^{-2}\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 32

A woodpecker’s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker’s head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm.

a) Find the acceleration in m/s2 and in multiples of g (g=9.80 m/s2).

b) Calculate the stopping time.

c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain’s deceleration, expressed in multiples of g?

Solution:

We are given the following values: v0=0.600 m/sv_0=0.600 \ \text{m/s}; vf=0.000 m/sv_f=0.000\:\text{m/s}; and Δx=0.002 m\Delta x=0.002\:\text{m}.

Part A

The acceleration is computed based on 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, we have

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=(0.000 m/s)2−(0.600 m/s)22(0.002 m)a=−90.0 m/s2  (Answer)\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0.000\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(0.002\:\text{m}\right)} \\ a & =-90.0\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

In terms of gg taking absolute values of the acceleration , we have

a=90.0 m/s2⋅(g9.80 m/s2)a=90.09.80ga=9.18g  (Answer)\begin{align*} a & = 90.0 \ \text{m/s}^2 \cdot \left( \frac{g}{9.80 \ \text{m/s}^2} \right) \\ a & = \frac{90.0}{9.80}g \\ a & = 9.18g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We shall use the formula

Δx=vavet\Delta x=v_{ave}t

where vavev_{ave} is the average velocity computed as

vave=v0+vf2v_{ave}=\frac{v_0+v_f}{2}

Solving for time tt in terms of the other variables, we have

t=2Δxv0+vft=\frac{2\Delta x}{v_0+v_f}

Substituting the given values, we have

t=2Δxv0+vft=2(0.002 m)0.600 m/s+0.000 m/st=6.67×10−3 s  (Answer)\begin{align*} t & =\frac{2\Delta x}{v_0+v_f} \\ t & =\frac{2\left(0.002\:\text{m}\right)}{0.600\:\text{m/s}+0.000\:\text{m/s}} \\ t & =6.67\times 10^{-3}\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

Employing the same formula we used in Part A, we have

a=(vf)2−(v0)22Δxa=(0.000 m/s)2−(0.600 m/s)22(0.0045 m)a=−40.0 m/s2\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0.000\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(0.0045\:\text{m}\right)} \\ a & =-40.0\:\text{m/s}^2 \end{align*}

In terms of gg, taking absolute values of the acceleration

a=40.0 m/s2⋅(g9.80 m/s2)a=40.09.80ga=4.08g  (Answer)\begin{align*} a & = 40.0 \ \text{m/s}^2 \cdot \left( \frac{g}{9.80 \ \text{m/s}^2} \right) \\ a & =\frac{40.0}{9.80}g \\ a & = 4.08g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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