A car starts moving along a line,first with accelaration a=2m/s^2,starting from rest then uniformally moving and finally deaccelarating at the...

Let the total distance travelled by the car be D. The
total time of motion is 10s; therefore the average speed is (D/10) m/s. We are given the
average speed as 3.2 m/s. So (D/10) = 3.2 or D= 32 m.

Now
the car starts from rest, accelerates for some time at 2 m/s^2, then moves at a uniform
speed and finally decelerates at 2m/s^2 to come to rest. As the car had started from
rest, the time for acceleration and deceleration is the same. Let the time the car
accelerates and decelerates be t1 and the time it moves at a uniform speed be t2. We
get: 2*t1+ t2 = 10.

Now the distance it travels while
accelerating is 2*t1*t1/2, this is the same as the distance it travels while
decelerating. The speed it has after t1 s is 2*t1. The distance travelled at the uniform
speed is 2*t1*t2.

This gives us 2*t1*t1/ 2 + 2*t1*t2 +
2*t1*t1/ 2 = 32.

Now 2*t1+ t2 =
10

=> t1 = (10 - t2)/
2

We substitute this in 2*t1*t1/ 2 + 2*t1*t2 + 2*t1*t1/ 2 =
32

=> 2*t1*t1 + 2*t1*t2 =
32

=> 2*[(10 - t2)/ 2] ^2 + 2*[(10 - t2)/ 2]*t2 =
32

=> [(10 - t2)/ 2] ^2 + [(10 - t2)/ 2]*t2 =
16

=> (10 - t2) ^2 + 2*10*t2 - 2*t2*t2 =
64

=> 100 + t2^2 - 2*10*t2 + 2*10*t2 - 2*t2*t2 =
64

=> 100 + t2^2 - 2*t2*t2 =
64

=> t2^2 =
36

=> t2= 6 or -6

As
time cannot be negative, so we have t2= 6
s

Therefore the car moves uniformly for 6
s