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:

We are given the following: v_0=0 \ \text{m/s}; v_f=26.8 \ \text{m/s}; and t=3.90\ \text{s}.

Part A

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

\begin{align*}
\overline{a} & =\frac{\Delta v}{\Delta t} \\
\overline{a} & =\frac{26.8\:\text{m/s}-0\:\text{m/s}}{3.90\:\text{s}} \\
\overline{a} & =6.872\:\text{m/s}^2\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

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

\begin{align*}
\Delta x & =v_{ave}t\\
\Delta x & =\left(\frac{0\:\text{m/s}+26.8\:\text{m/s}}{2}\right)\left(3.90\:\text{s}\right) \\
\Delta x & =52.26\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

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 x=v_{ave} t. That is,

\begin{align*}
\Delta x & = v_{ave} t \\
\Delta x & = \left(\frac{8\:\text{m/s}+40\:\text{m/s}}{2}\right)\left(3.33\times 10^{-2}\:\text{s}\right) \\
\Delta x & = 0.7992\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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

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, 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.

Part B

The list of known variables are:

Initial velocity: v_0=0\:\text{m/s}
Final Velocity: v_f=30.0\:\text{cm/s}
Distance Traveled: x-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 v_{ave} is the average velocity, and t is time. That is

\begin{align*}
\Delta x & =v_{ave} t \\
t &=\frac{\Delta x}{v_{ave}} \\
t & =\frac{1.80\:\text{cm}}{\frac{\left(0\:\text{cm/s}+30\:\text{cm/s}\right)}{2}}\\
t & =0.12\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part D

Since the computed value of the time for the 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.

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.