## Solution:

### Part A

The total distance traveled following Path D is

$d=\left(2\times 120\:m\right)+\left(6\times 120\:m\right)+\left(4\times 120\:m\right)+\left(1\times 120\:m\right)$

$d=1560\:m$

### Part B

Treat all motions upward and to the right positive, while downward and to the left negative.

The displacement in the x-direction is

$s_x=0+720\:m+0-120\:m=600\:m$

The displacement in the y-direction is

$s_y=-240\:m+0+480\:m+0=240\:m$

The total displacement is

$s=\sqrt{\left(s_x\right)^2+\left(s_y\right)^2}=\sqrt{\left(600\:m\right)^2+\left(240\:m\right)^2}=646.22\:m$

The direction angle from th x-axis is given by

$\theta _x=tan^{-1}\left|\frac{s_y}{s_x}\right|=tan^{-1}\left|\frac{240}{600}\right|=21.8^{\circ} \:North\:of\:East$

Therefore, the displacement is 646.22 m, directed at 21.8 degrees north of east.

## Solution:

### Part A

The total distance traveled following Path C is

$d=\left(1\times 120\:m\right)+\left(5\times 120\:m\right)+\left(2\times 120\:m\right)+\left(1\times 120\:m\right)+\left(1\times 120\:m\right)+\left(3\times 120\:m\right)$

$d=1560\:m$

### Part B

Treat all motions upward and to the right positive, while downward and to the left negative.

The displacement in the x-direction is

$s_x=0+600+0-120+0-360=120\:m$

The displacement in the y-direction is

$s_y=120+0-240+0+120+0=0\:m$

The total displacement is

$s=\sqrt{\left(s_x\right)^2+\left(s_y\right)^2}$

$s=\sqrt{\left(120\:m\right)^2+\left(0\:m\right)^2}$

$s=120\:m$

The direction angle from th x-axis is given by

$\theta _x=tan^{-1}\left|\frac{s_y}{s_x}\right|$

$\theta _x=tan^{-1}\left|\frac{0\:m}{120\:m}\right|$

$\theta _x=0^{\circ}$

Therefore, the displacement is 120 m, east.

## 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

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

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

$\Delta y=0.0051\:m$