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.

The Starship Enterprise returns from warp drive to ordinary space with a forward speed of 51 km/s. To the crew’s great surprise, a Klingon ship is 150 km directly ahead, traveling in the same direction at a mere 21 km/s. Without evasive action, the Enterprise will overtake and collide with the Klingons in just about 5.0 s. The Enterprise’s computers react instantly to brake the ship.

PART A. What magnitude acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship? Assume the acceleration is constant.

Solution:

Since the collision is barely avoided, we shall make their positions and their velocities equal. So, their velocities must be equal to the Klingon ship. For the Starship Enterprise, the final velocity will be equal to 21 km/s

Equate their positions as well

Substitute to all t’s

Therefore, the magnitude of the acceleration should be

A particle’s velocity is described by the function where is in m/s, t is in s, and k is a constant. The particle’s position at is . At , the particle is at .

PART A. Determine the units of k in terms of m and s.

ANSWER:

We know that the unit for velocity is m/s and the unit for time is s. Therefore,

A toy train is pushed forward and released at with a speed of 2.0 m/s. It rolls at a steady speed for 2.0 s, then one wheel begins to stick. The train comes to a stop 6.0 m from the point at which it was released.

What is the magnitude of the train’s acceleration after its wheel begins to stick? Assume acceleration is constant after wheel begins to stick.

ANSWER:

The toy train has a constant speed of from . Then, it began to decelerate at , and finally stop at . Considering the moment after the toy train’s wheel begin to stick up to the moment when it stopped.

The negative sign indicates that the acceleration is moving opposing the motion (a deceleration). Basically, the negative sign is just an indication of the direction of the acceleration. The magnitude of the acceleration is simple .

To understand the meaning of the variables that appear in the equations for one-dimensional kinematics with constant acceleration.

Motion with a constant, nonzero acceleration is not uncommon in the world around us. Falling (or thrown) objects and cars starting and stopping approximate this type of motion. It is also the type of motion most frequently involved in introductory kinematics problems.

The kinematic equations for such motion can be written as

where the symbols are defined as follows:

is the position of the particle;

is the initial position of the particle;

is the velocity of the particle;

is the initial velocity of the particle;

is the acceleration of the particle.

In answering the following questions, assume that the acceleration is constant and nonzero: a≠0.

PART A. The quantity represented by x is a function of time (i.e., is not constant).

ANSWER: True

PART B. The quantity represented by is a function of time (i.e., is not constant).

ANSWER: False

Recall that represents an initial value, not a variable. It refers to the position of an object at some initial moment.

PART C. The quantity represented by is a function of time (i.e., is not constant).

ANSWER: False

PART D. The quantity represented by v is a function of time (i.e., is not constant).

ANSWER: True

The velocity v always varies with time when the linear acceleration is nonzero.

PART E. Which of the given equations is not an explicit function of t and is therefore useful when you don’t know or don’t need the time?

ANSWER:

PART F. A particle moves with constant acceleration a. The expression vi+at represents the particle’s velocity at what instant in time?

ANSWER: when the time t has passed since the particle’s velocity was

More generally, the equations of motion can be written as

and

Here Δt is the time that has elapsed since the beginning of the particle’s motion, that is, , where is the current time and is the time at which we start measuring the particle’s motion. The terms and are, respectively, the position and velocity at . As you can now see, the equations given at the beginning of this problem correspond to the case , which is a convenient choice if there is only one particle of interest.

PART G. To illustrate the use of these more general equations, consider the motion of two particles, A and B. The position of particle A depends on time as . That is, particle A starts moving at time with velocity , from . At time , particle B has twice the acceleration, half the velocity, and the same position that particle A had at time t=0.

What is the equation describing the position of particle B?

ANSWER:

The general equation for the distance traveled by particle B is or , since is a good choice for B. From the information given, deduce the correct values of the constants that go into the equation for given here, in terms of A’s constants of motion.

PART H. At what time does the velocity of particle B equal that of particle A?

ANSWER:

Particle A’s velocity as a function of time is , and particle B’s velocity as a function of time is .

Once you have expressions for the velocities of A and B as functions of time, set them equal and find the time t at which this happens.