## Solution:

The velocity at time 2.5 seconds is 5 m/s.

The velocity at time 7.5 seconds is -4 m/s.

## Solution:

The position vs time graph is shown in the figure below.

## Solution:

Since the graph is a straight line, we can use the two points before and after the specified time to determine the slope of the line. The slope of the velocity-time graph is the acceleration.

The two points are $\left(0,\:165\right)\:and\:\left(20,\:228\right)$

The velocityis computed as

$a=m=\frac{\Delta y}{\Delta x}$

$a=\frac{y_2-y_2}{x_2-x_1}$

$a=\frac{228\:m/s-165\:m/s}{20\:s-0\:s}$

$a=3.2\:m/s^2$

Therefore, the acceleration is verified to be 3.2 m/s² at t=10 s.

## Solution:

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the position-time graph is the velocity.

The two points are $\left(20,\:6.95\right)\:and\:\left(40,\:11.7\right)$

The velocityis computed as

$v=m=\frac{\Delta y}{\Delta x}$

$v=\frac{y_2-y_2}{x_2-x_1}$

$v=\frac{11.7\:km-6.95\:km}{40\:s-20\:s}$

$v=0.238\:km/s$

Therefore, the velocity is 0.238 km/s.

## Solution:

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the position-time graph is the velocity.

The two points are $\left(0,\:2.80\right)\:and\:\left(20,\:6.95\right)$

The velocityis computed as

$v=m=\frac{\Delta y}{\Delta x}$

$v=\frac{y_2-y_2}{x_2-x_1}$

$v=\frac{6.95\:km-2.80\:km}{20\:s-0\:s}$

$v=0.208\:km/s$

Therefore, the velocity is 0.208 km/s.

## Solution:

### Part A

Based from the graph shown in figure 2.60, we can choose two points, one before and one after t=20 s. We have the points $\left(15,\:988\right)\:and\:\left(25,\:2138\right)$.

The slope is computed using the formula:

$m=\frac{\Delta y}{\Delta x}$

$m=\frac{2138\:m-988\:m}{25\:s-15\:s}$

$m=115\:m/s$

Therefore, we have verified that the velocity of the jet car is 115 m/s at t=20 seconds.

### Part B

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the velocity-time graph is the acceleration.

The two points are $\left(10,\:65\right)\:and\:\left(25,\:140\right)$

The acceleration is computed as

$a=m=\frac{\Delta y}{\Delta x}$

$a=\frac{y_2-y_2}{x_2-x_1}$

$a=\frac{140\:m/s-65\:m/s}{25\:s-10\:s}$

$a=5\:m/s^2$

Therefore, the acceleration is verified to be 5.0 m/s².

## Solution:

### Part A

This problem is the same as Problem 2.56. The velocity of the ball when it strikes the ground is

$v=\sqrt{\left(v_o\right)^2+2a\Delta y}$

$v=\sqrt{\left(0\:m/s\right)^2+2\left(-9.80\:m/s^2\right)\left(-1.50\:m\right)}$

$v=5.42\:m/s\:\left(downward\right)$

### Part B

The velocity of the ball just after it leaves the floor on its way back up

$v_o=\sqrt{v^2-2a\Delta y}$

$v_o=\sqrt{\left(0\:m/s\right)^2-2\left(-9.80\:m/s^2\right)\left(1.10\:m\right)}$

$v_o=4.64\:m/s$ upward

### Part C

The acceleration is

$a=\frac{v-v_o}{t}$

$a=\frac{4.643\:m/s-\left(-5.422\:m/s\right)}{0.0035\:s}$

$a=2880\:m/s^2$

### Part D

$\Delta y=\frac{\left(v_o\right)^2-v^2}{2a}$

$\Delta y=\frac{\left(-5.422\:m/s\right)^2-\left(0\:m/s\right)^2}{2\left(2880\:m/s^2\right)}$

$\Delta y=0.0051\:m$

## Solution:

### Part A

For the coin, we are given the following:  $v_o=+10\:m/s;\:a=-9.80\:m/s^2;\:v=0\:m/s\:\left(at\:top\:of\:ascent\right)$

The maximum height reached can be solved using the formula

$v^2=\left(v_o\right)^2+2a\Delta y$

Solve for Δy in terms of the other variables

$\Delta y=\frac{v^2-\left(v_o\right)^2}{2a}$

Now, substitute the given values

$\Delta y=\frac{\left(0\:m/s\right)^2-\left(10\:m/s\right)^2}{2\left(-9.80\:m/s\right)^2}$

$\Delta y=5.10\:m$

From the moment the coin is dropped to the top of the ascent, the change in position of the coin is +5.10 m. Therefore, the maximum height is

$y_{max}=300\:m+5.10\:m$

$y_{max}=305.10\:m$

### Part B

The position of the coin after 4 seconds can be solved using the formula

$\Delta y=v_ot+\frac{1}{2}at^2$

Substitute the given values

$y=\left(10\:m/s\right)\left(4\:s\right)+\frac{1}{2}\left(-9.80\:m/s^2\right)\left(4\:s\right)^2$

$y=-38.4\:m$

The change in position of the coin is -38.4 m. Therefore, the poisiton of the coin with respect to the ground is

$y_{_{\left(4\:s\right)}}=300\:m-38.4\:m$

$y_{_{\left(4\:s\right)}}=261.6\:m$

The velocity after 4 seconds is given by the formula

$v=v_o+at$

Substitute the given values

$v=10\:m/s+\left(-9.8\:m/s^2\right)\left(4.00\:s\right)$

$v=-29.2\:m/s$

The velocity is 29.2 m/s, downward.

### Part C

The time it takes for the coin to hit the ground can be solved using the formula

$\Delta y=v_ot+\frac{1}{2}at^2$

Substitute the given values

$300\:m=\left(10.0\:m/s\right)t+\frac{1}{2}\left(-9.80\:m/s^2\right)t^2$

Simplify and rearrange the equation to form a quadratic equation

$-4.90t^2+10t-300=0$

Now, we can solve for time, t, using the quadratic formula

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$t=\frac{-\left(10\right)\pm \sqrt{\left(10\right)^2-4\left(-4.9\right)\left(300\right)}}{2\left(-4.9\right)}$

$t=8.91\:s$

The time of flight is 8.91 seconds.

## Solution:

### Part A

We are given an initial velocity of 0. We are required to solve for the final velocity. We shall use the formula

$v^2=\left(v_o\right)^2+2a\Delta y$

Then, we substitute the given values

$v^2=\left(0\:m/s\right)^2+2\left(-9.80\:m/s^2\right)\left(0\:m-1.50\:m\right)$

$v=5.42\:m/s$

The velocity is 5.42 m/s downward.

### Part B

Consider the floor to be the initial position and the top of the trajectory to be the final position. We shall use the formula

$v^2=\left(v_o\right)^2+2a\Delta y$

The initial velocity is the unknown, so we solve the equation in terms of the initial velocity.

$v_o=\sqrt{v^2-2a\Delta y}$

Substitute the given values

$v_o=\sqrt{\left(0\:m/s\right)^2-2\left(-9.80\:m/s^2\right)\left(1.45\:m\right)}$

$v_o=5.33\:m/s$

The initial velocity is 5.33 m/s upward.

### Part C

The acceleration is calculated using the formula

$v=v_o+at$

Solve for the equation in terms of acceleration

$a=\frac{v-v_o}{t}$

Substitute the given values

$a=\frac{5.331\:m/s-\left(-5.422\:m/s\right)}{8.00\times 10^{-5}\:s}$

$a=1.34\times 10^5\:m/s^2$

### Part D

The period of compression occurs when the ball goes from -5.42 m/s to 5.33 m/s. The acceleration during this time is found in part C. The compression is solve using the formula

$v^2=\left(v_o\right)^2+2a\Delta y$

Solve for Δy in terms of the other variables

$\Delta y=\frac{\left(v^2\right)-\left(v_o\right)^2}{2a}$

Then we substitute the given values

$\Delta y=\frac{\left(5.42\:m/s\right)^2-\left(0\:m/s\right)^2}{2\left(1.34\times 10^5\:m/s^2\right)}$

$\Delta y=-1.09\times 10^{-4}\:m$

## Solution:

### Part A

We are required to solve for

$\vec{R'}=\vec{A'}+\vec{B}'$

The x-component of the resultant is:

$R_x'=A_x'+B_x'$

$R_x'=12\:cos\left(110^{\circ} \right)+\left(20.0\right)cos\left(40^{\circ} \right)=11.217\:m$

The y-component of the resultant is:

$R_x'=A_y'+B_y'$

$R_x'=\left(12.0\right)sin\left(110^{\circ} \right)+\left(20.0\right)sin\left(40^{\circ} \right)=24.132\:m$

The resultant is

$R'=\sqrt{R_x'+R_y'}$

$R'=\sqrt{\left(11.217\:m\right)^2+\left(24.132\:m\right)^2}$

$R'=26.6\:m$

The compass direction is

$\alpha =tan^{-1}\left(\frac{R_y'}{R_x'}\right)$

$\alpha =tan^{-1}\left(\frac{24.132}{11.217}\right)=65.1^{\circ}$

The compass direction is $65.1^{\circ} \:North\:of\:East$

### Part B

The result is the same as that in Part A since the components did not change. Only the order has changed. This is consistent with the fact that $\vec{A}-\vec{B}=-\left(\vec{B}-\vec{A}\right)$