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.

Solution:

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

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

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

The velocityis computed as

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

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.

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.

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.

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.

(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.

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 .

The slope is computed using the formula:

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

The acceleration is computed as

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

A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 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 0.0800 ms.

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

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

Then, we substitute the given values

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

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

Substitute the given values

The initial velocity is 5.33 m/s upward.

Part C

The acceleration is calculated using the formula

Solve for the equation in terms of acceleration

Substitute the given values

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

Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return.

(a) Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2.0000 s.

(b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 332.00 m/s in this well.

Solution:

Part A

So, we just need to solve for the distance y using the equation

Substitute the given values

Neglecting the speed of sound, the distance to the water is 19.6 m below the ground.

Part B

Let be the depth of the well, at the bottom of the well. Let be the speed of sound and be the time for the sound to travel from the bottom to the top of the well. Let be the time for the rock to reach the bottom.

Equation1

Equation2

Equation 3

Adding equation 2 and 3

Substitute the given values

Rearranging, we have

Solve the equation using the quadratic formula

We have solved for . Now, we can substitute this value to equation 2 to solve for

With the speed of sound taken into account, the distance to the water is about 18.5 meters.