a car started from rest and moves along straight road with constant acceleration of 5m/s2 for 8 seconds. what distance will the car cover in 12 sec since it started from rest?
Answers
Answer:
320 m
Explanation:
Distance moved in first 8 seconds can be calculated by using
s = ut + 1/2(a)(t)^2
where u = 0 since car started from rest
t = 8 seconds
a = 5 m/(s)^2
This comes out to be 160 m
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For rest of its motion car moves with the velocity it achieved in the course of 8 seconds due to the acceleration of 5 m/(s)^2
First we find the velocity at the end of 8 seconds using
v = u + at
where u = 0
a = 5
t =8
which comes out to be 40 m/s
Now distance moved in next four second of its motion is given by
s = vt
where v = 40
t = 4
which comes out to be 160 m
Therefore, total distance moved in 12 (8+4) seconds of motion is 320 m
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NOTE : As car moves on straight road and doesn't retards or changes its direction of motion displacement and distance practically mean one and the same thing.