A uniformly accelerated car passes three equally spaced traffic signs. The signs

Question

A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car passes the first sign at t = 1.2 s, the second sign at t = 3.9 s, and the third sign at t = 4.7 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? Partial credit: Your answer received partial credit. (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

d = (v1+v2)*(t2-t1)/2

V avg = v1+v2/2 = 25/(3.9-1.2)

      = 9.26 m/s

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b) d = (v2+v3)*(t3-t2)/2

V avg = v2+v3/2 = 25/(4.7-3.9) = 31.25 m/s

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a) from a & b

v1 + v2 = 18.52..............(1)

v2+v3 = 62.5...............(2)

2 - 1

v3 - v1 = 43.98

a = (v3-v1)/(t3-t1) = 12.56 m/s^2

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