A motor car starts from rest and accelerates uniformly for 10s to a velocity of 20 ms^–1. it then runs at a constant speed and is finally brought to rest in 40m with a constant acceleration. Total distance covered is 640 m. Find the value of acceleration retardation and total time taken.

Let us divide the question into three parts:

(1) Car starts from rest and accelerates to attain a velocity of 20 ms^–1 in time 10 s

(2) The car moves at a constant velocity

(3) Car decelerates and comes to rest after covering 40 m

The formula to be used:

..1

..2

..3

Case – 1:

Let acceleration = and distance =

v = 20 ms^–1; u = 0 ms^–1; t = = 10 s;

Now, using equation 1, we get:

..4

Now, using equation 2, we get:

..5

Case – 2:

Here the total distance that is given is = 640 m a, d also there are three parts.

So, let be distances travelled in the three parts.

From 5, we know that and also we know that.

So,

..6

Now, velocity = constant = 20 ms^–1

Now, for finding time:

..7

Case – 3:

Let acceleration = and time =

Now, we know that; v = 0 ms^–1; u = 20 ms^–1

So, using equation 3, we get:

..8

The sign of the acceleration is negative as it is for retardation.

Now, using equation 1, we get:

..9

So, = 4 s

Hence to answer the question:

(a) Acceleration = = (From 4)

(b) Retardation = = (From 8)

(c) Total Time Taken =

= 10 + 25 + 4 (From 7 & 9)

= 39 s