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

Question

A uniformly accelerated car passes three equally spaced trafficsigns. The signs are separated by a distanced=24 m. The car passes the first sign att1= 1.3 s, the secondsign att2= 3.9 s,and the third sign att3= 5.0 s.

1)What is the average velocity of the carduring the time it that it is moving between the first twosigns?

2)What is the average velocity of the car during the time that itpasses between the second and third signs?

3)What is the acceleration of the car?

I have the answers:

1)9.23

2)21.8

3)6.8

I understand #1 and 2 I need some one to explain #3

Explanation / Answer

a)Average velocity=displacement/time=24/(3.9-1.3)=9.23s

b)average velocity of the car during the time that itpasses between the second and third signs=displacement/time=24/(5-3.9)=21.8m/s

c)time interval for velocity change=(5+3.9)/2-(1.3+3.9)/2=1.85

What is the acceleration of the car=change in velocity/time=(21.8-9.23)/(1.85)=6.8m/s2

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