College Physics 2.29 – Acceleration of freight trains

Freight trains can produce only relatively small accelerations and decelerations.

(a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 minutes, starting with an initial velocity of 4.00 m/s?

(b) If the train can slow down at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity?

(c) How far will it travel in each case?


Part A

The final velocity can be solved using the formula \text{v}=\text{v}_0+\text{at}. We substitute the given values.


\displaystyle \text{v}=4.00\:\text{m/s}+\left(0.0500\:\text{m/s}^2\right)\left(8.00\:\text{min}\times \frac{60\:\sec }{1\:\min }\right)


Part B

Rearrange the equation we used in part (a) by solving in terms of t, we have

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

      \displaystyle \text{t}=\frac{0\:\text{m/s}-28\:\text{m/s}}{-0.550\:\text{m/s}^2}


Part C

The change in position for part (a), \Delta \text{x}, or distance traveled is computed using the formula \Delta \text{x}=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2

     \Delta \text{x}=\left(4.0\:\text{m/s}\right)\left(480\:\text{s}\right)+\frac{1}{2}\left(0.0500\:\text{m/s}^2\right)\left(480\:\text{s}\right)^2

     \Delta \text{x}=7680\:\text{m}

For the situation in part (b), the distance traveled is computed using the formula \Delta \text{x}=\frac{\text{v}^2-\text{v}_0^2}{2\text{a}}

     \displaystyle \Delta \text{x}=\frac{\left(0\:\text{m/s}\right)^2-\left(28.0\:\text{m/s}\right)^2}{2\left(-0.550\:\text{m/s}^2\right)}

     \Delta \text{x}=712.73\:\text{m}

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?


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.


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?


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}}


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?


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.




Part B

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




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.


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)

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) ?


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?


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.


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}