> > > > Problem 30E

# A cat walks in a straight line, which we shall call the ## Problem 30E Chapter 2

University Physics | 13th Edition

• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants University Physics | 13th Edition

4 5 0 405 Reviews
21
5
Problem 30E

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 this cat’s motion and construct a graph of the feline’s velocity as a function of time (?Fig. E2.30?). (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) Assuming that the cat started at the origin, sketch clear graphs of the cat’s acceleration and position as functions of time.

Step-by-Step Solution:

Solution 30E Step 1 of 6: (a) Find the cat’s velocity at t = 4.0 s and at t = 7.0 s. From the given v-t graph, Cat’s velocity at t= 4s is v= 2 cm/s Similarly at t= 7s, using straight line equation(y=mx+c), here v=mt-c Using slope,m= 1.33, c= 8 and t= 7s is v= -1.31 cm/s Step 2 of 6: (b) What is the cat’s acceleration at t = 3.0 s At t = 6.0 s At t = 7.0 s Cat’s acceleration can be calculated by the ratio of velocity by time, v a= t At t= 3 s, from graph v= 4 cm/s 2 2 Using 1 cm =10 m v= 4 × 10 m /s Using above equation, 4 ×10 m/s a= 3 s 2 2 a= 1.33 × 10 m/s At t= 6 s, from graph v= 0 cm/s Using above equation, a= 0 6 s a= 0 At t= 7 s, from graph v= -1.31 cm/s 2 2 Using 1 cm =10 m v= -1.31 × 10 m/s Using above equation, 1.31 ×10 m/s a= 3 s 2 2 a= - 0.43 × 10 m/s Step 3 of 6: (c) What distance does the cat move during the first 4.5 s From t = 0 to t = 7.5 s We know that, the area under the curve in the velocity -time graph will given the displacement. To calculate distance travelled for first 4.5 s Area= area of the trapezoid = a+bh 2 Using a=2, b=8 and h = 2.5 Displacement= area= 2+8 (2.5) 2 Displacement = 12.5 cm Therefore, the cat travels a distance of 12.5 cm in first 4.5 seconds.

Step 4 of 6

Step 5 of 6

##### ISBN: 9780321675460

Since the solution to 30E from 2 chapter was answered, more than 2718 students have viewed the full step-by-step answer. University Physics was written by and is associated to the ISBN: 9780321675460. This full solution covers the following key subjects: cat, velocity, acceleration, line, axis. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. The full step-by-step solution to problem: 30E from chapter: 2 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. The answer to “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 this cat’s motion and construct a graph of the feline’s velocity as a function of time (?Fig. E2.30?). (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) Assuming that the cat started at the origin, sketch clear graphs of the cat’s acceleration and position as functions of time.” is broken down into a number of easy to follow steps, and 126 words. This textbook survival guide was created for the textbook: University Physics, edition: 13.

#### Related chapters

Unlock Textbook Solution

A cat walks in a straight line, which we shall call the

×
Get Full Access to University Physics - 13 Edition - Chapter 2 - Problem 30e

Get Full Access to University Physics - 13 Edition - Chapter 2 - Problem 30e

I don't want to reset my password

Need help? Contact support

Need an Account? Is not associated with an account
We're here to help