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 = 25 m. The car passes the first sign at t = 1.3 s, the second sign at t = 3.7 s, and the third sign at t = 5.2 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

Given that , car travels in a distance, d=25 m

car passes a first sign is, t1=1.3 s

car passes a second sign is, t2=3.7 s

car passes a third sign is, t3=5.2 s

a)

average velocity = total distance / total time

Between the first two signs the total time is, t=t2-t1= 3.7 s - 1.3 s = 2.4 s

average velocity, V1= d/t=25 m/2.4 s=10.41 m/s

b)

average velocity = total distance / total time

Between the first two signs the total time is, t=t3-t2= 5.2 s - 3.7 s = 1.5 s

average velocity, V2= d/t=25 m/1.5 s=16.66 m/s

c)

average velocities of the first two signs are,

10.41 = (v2 + v1) / 2

v2 + v1 = 20.82 m/s ............(1)

16.66 = (v2 + v3) / 2
v2 + v3 = 33.32 m/s ............(2)

Subtracting two equations then

(v3+v2)-(v2+v1)=33.32 m/s-20.82 m/s

v3-v1=12.5 m/s

t3-t1=5.2 s-1.3 s=3.9 s

acceleration of the car is,

a = (v3 - v1) / (t3-t1)
So, a = 12.5 / 3.9 s = 3.2 m/s^2

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