## College Physics 3.14 – Distance and displacement of a path

#### In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition. Figure 3.58 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

## 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.

## College Physics 3.13 – Distance and Displacement

#### (b) the magnitude and direction of the displacement from start to finish.In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition. Figure 3.58 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

## 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 $\Delta y=\frac{\left(v_o\right)^2-v^2}{2a}$ $\Delta y=\frac{\left(-5.422\:m/s\right)^2-\left(0\:m/s\right)^2}{2\left(2880\:m/s^2\right)}$ $\Delta y=0.0051\:m$