Question

Maximum acceleration of a car is 4 m/s2 and
maximum retardation is 5 m/s2. The minimum time
in which this car travel a distance of 1 km is
(1) 40 s
(2) 60 S
(3) 30 s
(4) 20 s​

Answer 1

a is acceleration and d is deacceleration

speeds

initial speed is 0

then acceleration then v

then deaceleration then finally 0

displacement / time

initial to that point where acc. changes to deac. is v^2/2a

point where acc. changes to deac. to final is v^2/2d

(a is acceleration and d is deacceleration )

Total displacement is  1000m so v^2(9/40)

v=200/3

so total time would be

v/a+v/d=200/3(9/20)=30s

ANSWER: (3)-30s