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

Question

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 = 3.3 s, and the third sign at t = 5.3 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

Distance between sign = 22m

1st sign @ t = 1.2

2nd sign @ t = 3.3

3rd sign @ t = 5.3

a)

Average Velocity = Total displacement / Time taken

Total displacement between 1st two signs = 22m

Time taken = 3.3 - 1.2 = 2.1 s

Average Velocity = 22/2.1 m/s

Average Velocity = 10.48 m/s

b)

Average Velocity = Total displacement / Time taken

Total displacement between 2nd and 3rd sign = 22m

Time taken = 5.3 - 3.3 = 2 s

Average Velocity = 22/2 m/s

Average Velocity = 11 m/s

c)

Acceleration = Velocity / time

(v2 + v1) / 2 = 10.48
v2 + v1 = 20.84

(v2 + v3) / 2 = 11
v2 + v3 = 22

a = (v3 - v1) / (5.3 - 1.2)
a = (22 - 20.84 )/ 4.1
a = 0.283 m/s^s
Magnitude of the acceleration of the car a = 0.283 m/s^s

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