A car starts from rest and accelerates uniformly for 20 seconds to a velocity of 72 km`h^(–1)`. It then runs at constant velocity and finally brought to rest in 200 m with a constant retardation. The total distance covered is 600 m. Find the acceleration, retardation and the total time taken.

Explanation

(i) Motion with uniform acceleration
Here , u=0, `t_1=20` sec, `v=72xx5/(18)=20ms^(-1)`
`:.v=u+at_1`
`20=0+axx20` or `a=1ms^(-2)`
distance travelled by car in this time (20 sec)
`S_1=ut+1/2at^2=0+1/2xx1xx(20)^2=200m`
(ii) Motion with uniform velocity.
As given, total distance = 600 m we have calculated `S_1` = 200 m (with uniform acc.)
and `S_2` = 200 m (with retardation)
`:.` Net distance for which body moves with uniform velocity
`S=600-S_1-S_2=600-200-200=200m`
`:.` Time taken `t=("distance")/("uniform velociy")`
`=(200)/(20)=10sec`
`:.` Total time of journey, `t=(20+10+20)sec`
t=sec
Average velocity=`("Total displacement")/("Total Time")=(600)/(50)`
=12 m/s
(iii) for this motion, initial velocity, `u=20ms^(-1)` and
Final velocity `v=0,S_2=200m`
Acceleration a'=?
Using `v^2-u^2=2a'S_2`
`(20)^2=2(a')xx200`
a'=-1 `ms^(-2)`
Let t'= time for which the body comes to rest
`:.` v=u+a't
0=20-1t'
`:.` t'=20 sec