## Solution:

### Part A

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

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

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

$\text{v}=28.0\:\text{m/s}$

### Part B

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

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

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

$\text{t}=50.91\:\sec$

### 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 $\Delta \text{x}=\frac{\text{v}^2-\text{v}_0^2}{2\text{a}}$

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

$\Delta \text{x}=712.73\:\text{m}$

## 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

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