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A car starts from rest and moves along the x-axis with constant acceleration 5 m s–2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?

Answer:

Given,
Initial velocity, u=0,
acceleration, a=5 ms^{−2} and
time taken, t=8 s.

Let s1 be the distance travelled and v be the final velocity.

From second equation of motion, s=ut+1/2at^{2}
⇒s1=0×8+1/2×5×8^{2}=160 m

From first equation of motion, v=u+at⇒v=0+5×8=40 ms^{−1}

Now we have to find the distance covered in time t=12 s

We know, distance=speed×time.
As the speed is constant, after 8 s. i.e, speed =40 ms^{−1,} therefore in the remaining 4 s its covers distance, s2=40×4=160 m
Thus total distance covered in 12 s, s1+s2=160+160=320 m.

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