Question

A cat walks in a straight line
A cat walks in a straight line, which we shall call the x-axis with the positive direction to the right. As an observant physicist, you make measurements of thiscat's motion and construct a graph of the feline's velocity as a function of time (Fig. 2.36).
(a) Find the cat's velocity at t = 4.0 s and at t = 7.0 s.
(b) What is the cat's acceleration at t = 3.0 s? At t = 6.0 s? At t = 7.0 s?
(c) What distance does the cat move during the first 4.5 s? From t = 0 to t = 7.5 s?
(d) Sketch clear graphs of the cat's acceleration and position as functions of time, assuming that the cat started at the origin.

Answer 1

Concepts and reason

The concepts used to solve this problem are motion in a straight line and Newton’s second equation of motion.

Initially, use the given velocity–time graph to find the velocity.

Then, use slope of the graph to find the acceleration.

Then, use Newton’s second equation of motion to find the distance.

Finally, use the velocity–time graph to plot the acceleration–time graph.

Fundamentals

The acceleration is the rate of change of velocity of an object with respect to time.

Here, is the acceleration, is the velocity, and is the time.

The expression for Newton’s second equation of motion is as follows:

Here, is the final position, is the initial position, and is the initial velocity of the moving object.

The relation between initial and final velocity is as follows:

Here, is the final velocity.

(a)

The acceleration is the rate of change of velocity of an object with respect to time.

$=D $

From the given graph, substitute for and .

The relation between initial and final velocity is as follows:

$v=u+at $

Here, velocity at $t=4.0s $ is .

The velocity at $t=4.0s $ is found from the given graph:

Substitute for , for , and for .

The expression for velocity at is as follows:

$V2=u+at $

Here, velocity at is .

Substitute for , for , and for .

(b)

The expression for the acceleration is given below:

The given graph is the velocity–time graph. Hence, slope of the given graph gives the acceleration, and the acceleration at any time will be equal for this motion.

From the given graph:

Substitute for and .

(c)

From to $t=4.5s $,

Substitute for, for , and for .

From to ,

Substitute for, for and for .

(d)

Since the velocity is uniform, the acceleration from the velocity–time graph is a constant. The acceleration–time graph of the motion is given below:

Ans: Part a

Thus, the velocities at $t=4.0s $ and $t=7.0s $ are and respectively.

Part b

Thus, the acceleration is .

Part c

Thus, the distance that the cat moves during the first is and from $t=0s $ to is .

Part d

Thus, the acceleration–time graph is given below: