Tag Archives: motion in one dimension

College Physics by Openstax Chapter 2 Problem 25

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:

We are given the following: v_0=9.00 \ \text{m/s}; and a=2.00 \ \text{m/s}^2.

Part A

For this part, we are given t=5.00 \ \text{s} and we shall use the formula  x=x_0+v_0 t+\frac{1}{2}at^2.

\begin{align*}
x & =x_0+v_0 t+\frac{1}{2}at^2 \\
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 \\
x & =20\:\text{meters} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

Part B

The final velocity can be determined using the formula v_f=v_0+at.

\begin{align*}
v_f & =v_0+at \\
v_f & =9.00\:\text{m/s}+\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right) \\
v_f & =-1\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

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.

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College Physics by Openstax Chapter 2 Problem 24

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s.

(a) Draw a sketch of the situation.

(b) List the knowns in this problem.

(c) How far does the car travel in those 12.0 s? 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, check your units, and discuss whether the answer is reasonable.

(d) What is the car’s final velocity? Solve for this unknown in the same manner as in part (c), showing all steps explicitly.

Solution:

Part A

The sketch of the situation is shown below. Also, the knowns and unknowns are in the illustration.

College Physics Problem 2.24 Illustration

From the illustration above, we can see that the initial velocity is 0 m/s, the initial time and initial distance are also zero. The final velocity and the final distance are unknowns. The time at the final location is 12 seconds and the acceleration is constant all throughout the trip at 2.40 meters per second square.

Part B

The knowns are: a=2.40\:\text{m/s}^2; t=12.0\:\sec; v_0=0\:\text{m/s}; and x_0=0\:\text{m}

Part C

For this part, the unknown is the value of x. If we examine the equations for constant acceleration and the given values in this problem, we can readily use the formula x=x_0+v_0 t+\frac{1}{2}at ^2. That is

\begin{align*}
x & =x_0+v_0 t+\frac{1}{2}at^2 \\
x & =0\:\text{m}+\left(0\:\text{m/s}\right)\left(12.0\:\text{s}\right)+\frac{1}{2}\left(2.40\:\text{m/s}^2\right)\left(12.0\:\text{s}\right)^2 \\
x & =172.8\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

Part D

For this part, the unknown is the value of v. The equation that can be used based from the known variables is v=v_0+at. That is

\begin{align*}
v & =v_0+at \\
v & =0\:\text{m/s}+\left(2.40\:\text{m/s}^2\right)\left(12.0\:\text{s}\right) \\
v & =28.8\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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