8. Use the exact values you enter to make later calculations. A uniformly accele

Question

8. Use the exact values you enter to make later calculations.

A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 22 m. The car passes the first sign at t = 1.2 s, the second sign at t = 3.2 s, and the third sign at t = 5.0 s.

(a) What is the magnitude of the average velocity of the car during the time that it is moving between the first two signs?

(b) What is the magnitude of the average velocity of the car during the time that it is moving between the second and third signs?

(c) What is the magnitude of the acceleration of the car?

Explanation / Answer

Vavg = total displacement / total time

a) here total displacement = 22m

time taken = 1.2 sec

Vavg = 22/1.2 = 18.33m/sec

b)

here total displacement = 22m

time taken = 3.2-1.2 = 2sec

Vavg = 22/2 = 11 m/sec

c)

22 = v(1.2) +0.5*a*1.2*1.2

44 = v (3.2) +0.5*a*3.2*3.2

from above equations we get

a = 18.33 m/sec^2

Related Questions

Navigate

• Browse (All)
• Browse 8
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.