Question

a car start from rest and move along the x axis with constant accelaration of 5 m/sec2 for 8 second. if it then continue with same acceleration, what distance will the car cover in 12 second since it started from rest

Answer 1

Answer:

Here, u=0,a=5m/s2,t=8su=0,a=5m/s2,t=8s

From s=ut+12at2s=ut+12at2

s=0+12×5×82=160ms=0+12×5×82=160m

This is the distance moved by the car is first 8sec8sec. Velocity at the end of 8s8s is

v=u+at=0+5×8=40m/sv=u+at=0+5×8=40m/s.

As this velocity is uniform, distance moved in next 4s4s

s'=v×t=40×4=160ms′=v×t=40×4=160m

Total distance covered by the car in 12s12s

=s+s'=160+160=320m

Explanation:

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Answer 2

Answer:

320 m

Explanation:

We know that,

S = u + ½ at²

The initial velocity is zero. Therefore, the distance covered by the car in first 8 seconds is

S = 0 + ½ × 5 × 8²

S = 160 meters

Therefore, the velocity attained at the end of 8 seconds is v = u + at

v = 0+ 5×8

v = 40 m/s

since the car has moved with constant velocity from 8th second to 12 second, the distance traveled is

S = v× t = 40 × 4 = 160 m

Therefore, the total distance traveled is 160+160 = 320 m

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