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College Physics by Openstax Chapter 2 Problem 46


A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool.

(a) How long are her feet in the air?

(b) What is her highest point above the board?

(c) What is her velocity when her feet hit the water?


Solution:

The known values are: y_0=1.80\:\text{m}, y=0\:\text{m}, a=-9.80\:\text{m/s}^2, v_0=4.00\:\text{m/s}.

Part A

Based from the knowns, the formula most applicable to solve for the time is \Delta y=v_0t+\frac{1}{2}at^2. If we rearrange the formula by solving for t, and substitute the given values, we have

\begin{align*}
t & =\frac{-v_0\pm \sqrt{v_0^2-2a\Delta y}}{a} \\
t & =\frac{-4.00\:\text{m/s}\pm \sqrt{\left(4.00\:\text{m/s}\right)^2-2\left(-9.80\:\text{m/s}^2\right)\left(1.80\:\text{m}\right)}}{-9.80\:\text{m/s}^2} \\
t & =1.14\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We have the formula

\begin{align*}
\Delta y & =\frac{v^2-v_0^2}{2a} \\
\Delta y & =\frac{\left(0\:\text{m/s}\right)^2-\left(4.00\:\text{m/s}\right)^2}{2\left(-9.80\:\text{m/s}^2\right)} \\
\Delta y & =0.816\: \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

The formula to be used is

v^2=v_0^2+2a\Delta y

Substituting the given values

\begin{align*}
v^2 & =v_0^2+2a\Delta y \\
v & =\pm \sqrt{v_0^2+2a\Delta y} \\
v & =\pm \sqrt{\left(4.00\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(-1.80\:\text{m}\right)} \\
v & =\pm \sqrt{51.28\:\text{m}^2/\text{s}^2} \\
v &=\pm 7.16\:\text{m/s}
\end{align*}

Since the diver must be moving in the negative direction,

v=-7.16\:\text{m/s}  \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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College Physics by Openstax Chapter 2 Problem 45


A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.(a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable.(c) How long is the dolphin in the air? Neglect any effects due to his size or orientation.


Solution:

We will treat the downward direction as negative, and the upward direction as positive.

Part A

The known values are:a=-9.80\:\text{m/s}^2; v_0=13\:\text{m/s}; and y_0=0\:\text{m}.

Part B

At the highest point of the jump, the velocity is equal to 0. For this part, we will treat the initial position at the moment it jumps out of the water, and the final position at the highest point. Therefore, v_f=0 \text{m/s}.

The unknown is the final position, y_f. We are going to use the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta y \\
\text{or} \\
\left(v_f\right)^2=\left(v_0\right)^2+2a\left(y_f-y_0\right)

Solving for y_f in terms of the other variables:

y_f=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a}+y_0

Substituting the given values:

\begin{align*}
y_f & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a}+y_0 \\
y_f & =\frac{\left(0\:\text{m/s}\right)^2-\left(13.0\:\text{m/s}\right)^2}{2\left(-9.80\:\text{m/s}^2\right)}+0\:\text{m} \\
y_f & =8.62\:\text{m}+0\:\text{m} \\
y_f & =8.62\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

This value is reasonable since dolphins can jump several meters high out of the water. Usually, a dolphin measures about 2 meters and they can jump several times their length.

Part C

The unknown is time, \Delta t. We are going to use the formula

v_f=v_0+at

Solving for time, \Delta t in terms of the other variables:

t=\frac{v_f-v_0}{a}

Substituting the given values:

\begin{align*}
t & =\frac{v_f-v_0}{a} \\
t & =\frac{0\:\text{m/s}-13.0\:\text{m/s}}{-9.80\:\text{m/s}^2} \\
t &=1.3625\:\text{s}
\end{align*}

This value is the time it takes the dolphin to reach the highest point. Since the time it takes to reach this point is equal to the time it takes to go back to the water, the time it is in the air is:

\begin{align*}
t_{air} & =2\times t \\
t_{air}&=2\times 1.3625\:\text{s} \\
t_{air}&=2.65\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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College Physics by Openstax Chapter 2 Problem 44


A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.


Solution:

We will treat the upward direction as positive, and the downward direction as negative.

Part A

The known values are: a=-9.80 \text{m/s}^2; v_0=-1.40\:\text{m/s}; \Delta t=1.8\:\text{s}; and y_f=0\:\text{m}

Part B

We are looking for the initial position, y_0. We are going to use the formula

\Delta y=v_{0y}t+\frac{1}{2}at^2 
\\
\text{or}
\\
y_f-y_0=v_{0y}t+\frac{1}{2}at^2

Solving for y_0 in terms of the other variables:

y_0=y_f-v_{0y}t-\frac{1}{2}at^2

Substituting the given values:

\begin{align*}
y_0 & =y_f-v_{0y}t-\frac{1}{2}at^2 \\
y_0& =0-\left(-1.4\:\text{m/s}\right)\left(1.8\:\text{s}\right)-\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(1.8\:\text{s}\right)^2 \\
y_0&= 0-\left(-1.4\:\text{m/s}\right)\left(1.8\:\text{s}\right)-\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(1.8\:\text{s}\right)^2 \\ 
y_0& = 0+2.52\:\text{m}+15.876\:\text{m} \\
y_0& =18.396\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

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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: v_{fy}=0 \ \text{m/s}; \Delta y=1.25 \ \text{m}; and a=-9.80 \ \text{m/s}^2.

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

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

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

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

Substituting the given values:

\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 42


Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.


Solution:

The given known quantities are: a=-9.8\:\text{m/s}^2; y_0=0 \ \text{m}; and v_0=-14 \ \text{m/s}.

To compute for the displacement, we use the formula

\Delta y=v_0t+\frac{1}{2}at^2

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

v_f=v_0+at

Part A

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

\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\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 & =-8.23\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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

\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right) \\
& =-18.9\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The displacement at t=1.00\ \text{s} is

\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\text{m/s}\right)\left(1.00\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.00\:\text{s}\right)^2 \\
\Delta y & =-18.9\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The velocity at t=1.00\ \text{s} is

\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.00\:\text{s}\right) \\
& =-23.8\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

The displacement at t=1.50\ \text{s} is

\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\text{m/s}\right)\left(1.50\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.50\:\text{s}\right)^2 \\
\Delta y & =-32.0\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The velocity at t=1.50\ \text{s} is

\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.50\:\text{s}\right) \\
& =-28.7\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part D

The displacement at t=2.00\ \text{s} is

\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\text{m/s}\right)\left(2.00\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.00\:\text{s}\right)^2 \\
\Delta y & =-47.6\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The velocity at t= 2.00 \ \text{s} is

\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.00\:\text{s}\right) \\
& =-33.6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part E

The displacement at t=2.50\ \text{s} is

\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\text{m/s}\right)\left(2.50\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.50\:\text{s}\right)^2 \\
\Delta y & =-65.6\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The velocity at t= 2.50 \ \text{s} is

\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.50\:\text{s}\right) \\
& =-38.5\:\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\:\text{m/s}^2; y_o=0\:\text{m}; and v_{oy}=+15\:\text{m/s}.

To compute for the displacement, we use the formula

\Delta y=v_{oy}t+\frac{1}{2}at^2

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

v_{fy}=v_{oy}+at

Part A

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

\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 \ \text{s} is

\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 \ \text{s} is

\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\ \text{s} is

\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\ \text{s} is

\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\ \text{s} is

\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\ \text{s} is

\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\ \text{s} is

\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 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 \Delta x_1 (while accelerating) and \Delta x_2 (with constant speed), where \Delta x_1 + \Delta x_2 = 100 \ \text{m} .

For \Delta x_1:

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

\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 \ \text{s}, so

\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 \Delta x_1 + \Delta x_2 = 100 \ \text{m} .

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

Plugging in the given values

\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:

\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: v_0=0 \ \text{mph} ; v_f=60 \ \text{mph}; and \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=\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

v_f=v_0+at

Solving for time t in terms of the other variables

t=\frac{v_f-v_0}{a}

Substituting the given values to solve for t_1, the time it takes to accelerate from rest to maximum velocity:

\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 \Delta x_1 during this period is computed using the formula

\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:

\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: \Delta x_2=5.0\:\text{mi}-0.3\:\text{mi}=4.7\:\text{mi} .

We are going to solve t_2, the time spent on the course at max speed using the formula

\Delta x_2=v_{max}t_2

Solving for t_2 in terms of the other variables:

t_2=\frac{\Delta x_2}{v_{max}}

Substituting the given values:

\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

\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: v_0=11.5 \ \text{m/s} ; a=0.500 \ \text{m/s}^2; and \Delta t=7.00 \ \text{s}.

Part A

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

v_f=v_0+at

Substituting the given values:

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

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

\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' be the time it will take the rider at the constant final velocity to complete the race:

\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 T it will take the accelerating rider to reach the finish line is 

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

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

\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: \Delta x'=295 \ \text{m} ; v'=11.8 \ \text{m/s}

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

We are going to use the formula

t_2=\frac{\Delta x'}{v'}

Substituting the given values:

\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

\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

\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: v_0=0\ \text{m/s} ; v_f=145 \ \text{m/s}; and \Delta t=4.45 \ \text{sec} .

Part A

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

a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_0}{\Delta t}

Substituting the given values, we have

\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 \ \text{m/s}^2 ; v_0=0 \ \text{m/s}; and \Delta x=402 \ \text{m} .

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

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

To compute for the final velocity, we have

v_f=\sqrt{\left(v_0\right)^2+2a\Delta \:x}

Substituting the given values:

\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\:\text{m/s}.


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