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 = 24 m. The car passes the first sign at t = 1.5 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? 10 m/s Correct: Your answer is correct. (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? 30 m/s Correct: Your answer is correct. (c) What is the magnitude of the acceleration of the car?

Explanation / Answer

Here ,

distance , d = 24m

at t1 = 1.5 s , t2 = 3.9 s , t3 = 4.7 s

a) average velocity during the time = distance/time

average velocity during the time = 24/(2.9 -1.5) m/s

average velocity during the time = 10 m/s

the average velocity during the time is 10 m/s

b) between the second and third signs

average velocity during the time = 24/(4.7 - 3.9) m/s

average velocity during the time = 30 m/s

the average velocity during the time is 30 m/s

c)

let the magnitude of acceleration is a = (v^2 - u^2)/(2 *d)

the magnitude of acceleration is a = (30^2 - 10^2)/(2 * 24)

the magnitude of acceleration is a = 16.7 m/s^2

the the magnitude of acceleration is 16.7 m/s^2

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