Question

A car accelerates from rest at a constant rate of 3 metre per second square for some time. Then it retards at constant rate of 6 m per second square and comes to rest. If the total time for which it remains in motion is 3 seconds, what is the total distance travelled?

Answer 1

Dear student,
first we need to find the time it accelerated and retarded.
Let the time of acceleration be t and that of retardation be (3-t).(it is given total time is 3s).
Let the final velocity attained after acceleration be v.
By a = v-u/t
here u=0( it starts from rest)
We get 3 = v - 0/t
=V=3t.
Now before retardation the speed was v( as retardation started after acceleration). So for retardation u=v(acceleration)=3t.
Now final velocity after retardation is 0 as it is at rest. So we have:
-6= 0-u/(3-t)
Minus sign is to denote retardation.
-u = -18+6t
-3t = -18 +6t
​​​​​18= 9t
t=2s.


So time of acceleration is 2s. And that of retardation is 3-2= 1s.
So distance travelled during acceleration is

S= ut+at²/2
S= 0+ 3×2²/2
S=6m.

Now during retardation:
S= ut +at²/2
S= (3t)×1 + -6 ×1²/2 (see previous part carefully)

S= 6 - 6/2
S= 3m.
Total distance = 6+3 =9m.
Thank you.