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

## 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, $\Delta \text{x}=\text{v}_{\text{ave}}\text{t}$ $\displaystyle \Delta \text{x}=\left(\frac{0\:\text{m/s}+26.8\:\text{m/s}}{2}\right)\left(3.90\:\text{s}\right)$ $\Delta \text{x}=52.26\:\text{m}$

## 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.

## 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: $\text{v}_0=0\:\text{m/s}$
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.

## 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.

## 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

The values are plotted in the coordinate axes

## 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.

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

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

## Solution:

The formula for acceleration is $\displaystyle \text{a}=\frac{\text{change in velocity}}{\text{time}}=\frac{\Delta \text{v}}{\text{time}}=\frac{\text{v}_{\text{f}}-\text{v}_0}{\text{t}}$

Substituting the given values $\displaystyle \text{a}=\frac{30.0\:\text{m/s}-0\:\text{m/s}}{7.00\:\text{s}}=4.29\:\text{m/s}^2$