College Physics 2.65 – Velocity graph of a world-class sprinter


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 for 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

\text{v}_{\text{ave}}=6\:\text{m/s}

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

\text{a}=\text{slope}=\frac{12\:\text{m/s}-0\:\text{m/s}}{4\:\text{s}}=3\:\text{m/s}^2

Part D

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

\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

\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

\text{t}_{\text{total}}=4\:\sec +6.3\:\sec =10.3\:\sec


College Physics 2.28 – Acceleration of a powerful motorcycle


A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s.

(a) What is its average acceleration?

(b) How far does it travel in that time?


Solution:

Part A

The average acceleration of the motorcycle can be solved using the equation \overline{\text{a}}=\frac{\Delta \text{v}}{\Delta \text{t}}. Substitute the given into the equation. That is,

\displaystyle \overline{\text{a}}=\frac{\Delta \text{v}}{\Delta \text{t}}

\displaystyle \overline{\text{a}}=\frac{26.8\:\text{m/s}-0\:\text{m/s}}{3.90\:\text{s}}

\displaystyle \overline{\text{a}}=6.872\:\text{m/s}^2

Part B

The distance traveled is equal to the average velocity multiplied by the time of travel. That is,

\Delta \text{x}=\text{v}_{\text{ave}}\text{t}

\displaystyle \Delta \text{x}=\left(\frac{0\:\text{m/s}+26.8\:\text{m/s}}{2}\right)\left(3.90\:\text{s}\right)

\Delta \text{x}=52.26\:\text{m}


College Physics 2.27 – Acceleration of a slap shot


In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes 3.33×10-2 s, calculate the distance over which the puck accelerates.


Solution:

The best equation that can be used to solve this problem is \Delta \text{x}=\text{v}_{\text{ave}}\text{t}. That is,

\Delta \text{x}=\text{v}_{\text{ave}}\text{t}

\displaystyle \Delta \text{x}=\left(\frac{8\:\text{m/s}+40\:\text{m/s}}{2}\right)\left(3.33\times 10^{-2}\:\text{s}\right)

\Delta \text{x}=0.7992\:\text{m}

Therefore, the distance over which the puck accelerates is 0.7992 meters.


College Physics 2.26 – Blood acceleration


Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart.

(a) Make a sketch of the solution.

(b) List the knowns in this problem.

(c) How long does the acceleration take? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units.

(d) Is the answer reasonable when compared with the time for a heartbeat?


Solution:

Part A

The sketch should contain the starting point and the final point. This will be done by connecting a straight line from the starting point to the final point. The sketch is shown below.

College Physics Problem 2.26

Part B

The list of known variables are:

Initial velocity: \text{v}_0=0\:\text{m/s}
Final Velocity: \text{v}=30.0\:\text{cm/s}
Distance Traveled: \text{x}-\text{x}_0=1.80\:\text{cm}

Part C

The best equation to solve for this is \:\:\Delta \text{x}=\text{v}_{\text{ave}}\text{t} where \text{v}_{\text{ave}} is the average velocity, and t is time. That is

\Delta \text{x}=\text{v}_{\text{ave}}\text{t}

\displaystyle \:\text{t}=\frac{\Delta \text{x}}{\text{v}_{\text{ave}}}

\displaystyle \:\text{t}=\frac{1.80\:\text{cm}}{\frac{\left(0\:\text{cm/s}+30\:\text{cm/s}\right)}{2}}

\:\:\:\text{t}=0.12\:\text{s}

Part D

Since the computed value of the time for acceleration of blood out of the ventricle is only 0.12 seconds (only a fraction of a second), the answer seems reasonable. This is due to the fact that an entire heartbeat cycle takes about one second. So, the answer is yes, the answer is reasonable.


College Physics 2.25 – Deceleration of a race runner


At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of 2.00 m/s2.

(a) How far does she travel in the next 5.00 s?

(b) What is her final velocity?

(c) Evaluate the result. Does it make sense?


Solution:

Part A

For this part, we use the formula \text{x}=\text{x}_0\text{t}+\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2.

\text{x}=\text{x}_0\text{t}+\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2

\text{x}=0\:\text{m}+\left(9.00\:\text{m/s}\right)\left(5.00\:\text{s}\right)+\frac{1}{2}\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right)^2

\text{x}=20\:\text{meters}

Part B

The final velocity can be determined using the formula \text{v}=\text{v}_0+\text{at}

\text{v}=\text{v}_0+\text{at}

\text{v}=9.00\:\text{m/s}+\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right)

\text{v}=-1\:\text{m/s}

Part C

The result says that the runner starts at the rate of 9 m/s and decelerates at 2 m/s2. After some time, the velocity is already negative. This does not make sense because if the velocity is negative, that means that the runner is already running backwards.


College Physics 2.20 – Acceleration of an Olympic-class sprinter


An Olympic-class sprinter starts a race with an acceleration of 4.50 m/s2.

(a) What is her speed 2.40 s later?

(b) Sketch a graph of her position vs. time for this period.


Solution:

We are given \displaystyle \text{a}=4.50\:\text{m/s}^2,\:\text{t}=2.40\:\sec ,\:\text{and}\:\text{v}_0=0\:\text{m/s}

Part A

The unknown is vf. The formula in solving for vf is

\displaystyle \text{v}_{\text{f}}=\text{v}_0+\text{at}

Substituting the given values,

\displaystyle \text{v}_{\text{f}}=0\:\text{m/s}+\left(4.50\:\text{m/s}^2\right)\left(2.40\:\text{s}\right)=108\:\text{m/s}

Part B

The relationship between position and time can be calculated using the formula

\displaystyle \text{x}=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2

Then, with the given, we can express position in terms of time

\displaystyle \text{x}=0+\frac{1}{2}\left(4.50\:\text{m/s}^2\right)\left(\text{t}^2\right)=2.52\text{t}^2

The values of the position given the time is tabulated below

Time (s)Position (m)
00
12.25
29
320.25
436
556.25
681
7110.25
8144
9182.25
10225

The values are plotted in the coordinate axes 

Time vs Position

College Physics 2.19 – Acceleration of an intercontinental ballistic missile


Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in m/s2 and in multiples of g (9.80 m/s2) ?


Solution:

The formula for acceleration is 

\displaystyle \text{a}=\frac{\text{v}_{\text{f}}-\text{v}_0}{\text{t}}

Substituting the given values

\displaystyle \text{a}=\frac{6.5\times 10^3\:\text{m/s}-0\:\text{m/s}}{60.0\:\text{sec}}=108.33\:\text{m/s}^2

This can be expressed in multiples of g

\displaystyle \text{a}=\frac{108.33\:\text{m/s}^2}{9.80\:\text{m/s}^2}=11.05\text{g}

Therefore, the average acceleration is 108.33 m/s2 and can be expressed as 11.05g.


College Physics 2.18 – Speed of a car backing out of the garage


A commuter backs her car out of her garage with an acceleration of 1.40 m/s2 .

(a) How long does it take her to reach a speed of 2.00 m/s?

(b) If she then brakes to a stop in 0.800 s, what is her deceleration?


Solution:

Part A

The formula for acceleration is 

\displaystyle \text{a}=\frac{\text{v}_{\text{f}}-\text{v}_0}{\text{t}}

If we rearrange the formula by solving for the t, in terms of velocity and acceleration, we come up with

\displaystyle \text{t}=\frac{\text{v}_{\text{f}}-\text{v}_0}{\text{a}}

Substituting the given values, we have

\displaystyle \text{t}=\frac{2.00\:\text{m/s}-0\:\text{m/s}}{1.40\:\text{m/s}^2}=1.43\:\text{seconds}

Part B

The formula for acceleration (deceleration) is 

\displaystyle \text{a}=\frac{\text{v}_{\text{f}}-\text{v}_0}{\text{t}}

Then substituting all the given values, we have

\displaystyle \text{a}=\frac{0\:\text{m/s}-2\:\text{m/s}}{0.8\:\text{m/s}^2}=-2.50\:\text{m/s}^2


College Physics 2.17 – Extreme Deceleration on the Human Body


Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his

(a) acceleration and

(b) deceleration.

Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity.


Solution:

Part A

The formula for acceleration is 

\displaystyle \text{a}=\frac{\text{v}_{\text{f}}-\text{v}_{\text{0}}}{\text{t}}

Substituting the given values

\displaystyle \text{a}=\frac{282\:\text{m/s}-0\:\text{m/s}}{5.00\:\sec }=56.4\:\text{m/s}^2

Part B

The deceleration is 

\displaystyle \text{a}=\frac{0\:\text{m/s}-282\:\text{m/s}}{1.40\:\text{s}}=-201.43\:\text{m/s}^2

In expressing the computed values in terms of g, we just divide them by 9.80.

The acceleration is 

\displaystyle \frac{56.4}{9.80}=5.76\text{g}

The deceleration is 

\displaystyle \frac{201.43}{9.80}=20.55\text{g}