College Physics 3.14 – Distance and displacement of a path


Find the following for path D in Figure 3.58:

(a) the total distance traveled and

(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.

3.1
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


Find the following for path C in Figure 3.58:

(a) the total distance traveled and

(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.

The figure shows C starts from a point, then goes up 1 block, then 5 blocks to the right, then 2 blocks downward, 1 block to the left, 1 block upward, then 3 blocks to the left
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. 


Vector Addition and Subtraction|College Physics| Problem 3.5


Suppose you first walk 12.0 m in a direction 20º west of north and then 20.0 m in a direction 40.0º south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , as in Figure 3.56, then this problem finds their sum R = A + B .)

3.5

Continue reading “Vector Addition and Subtraction|College Physics| Problem 3.5”

College Physics 2.63 – Constructing a displacement graph


Construct the displacement graph for the subway shuttle train as shown in Figure 2.18(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

2.63


Solution:

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

2.63b

College Physics 2.62 – Acceleration as the slope of the velocity vs time diagram


By taking the slope of the curve in Figure 2.63, verify that the acceleration is 3.2 m/s²  at t = 10 s.

2.62
Figure 2.63

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.


College Physics 2.61 – Velocity as slope of position vs time


Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t = 30.0 s is 0.238 m/s. Assume all values are known to 3 significant figures.

2.60
Figure 2.62

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.


College Physics 2.60 – Slope of the position vs time diagram


Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t = 10.0 s is 0.208 km/s. Assume all values are known to 3 significant figures.

2.60
Figure 2.62

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.


College Physics 2.59 – Analysis of motion diagrams


(a) By taking the slope of the curve in Figure 2.60, verify that the velocity of the jet car is 115 m/s at t = 20 s.

(b) By taking the slope of the curve at any point in Figure 2.61, verify that the jet car’s acceleration is 5.0 m/s².

Figure 2.60
Figure 2.61

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.


College Physics 2.58 – A tennis ball dropped onto a hard floor


A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m.

(a) Calculate its velocity just before it strikes the floor.

(b) Calculate its velocity just after it leaves the floor on its way back up.

(c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (0.0035 s).

(d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?


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