For the car to cover the distance in minimum time, it must continue to accelerate for as long as possible. If the car first constantly accelerates, and then immediately constantly decelerates, then the minimum time is achieved.
Firstly, we need to calculate the maximum acceleration of the car. We know that the car reaches the speed of 162 km/h after 5 s of acceleration. Let's assume v to be the final speed of the car.
Then,
v=162 km/h=162×518=45 m/s
So we know,
v=45 m/s
u=0 m/s
t=5 s
By using v=u+at we get,
⇒45=0+a×5
⇒a=9 m/s2
Let's assume that to cover 1 km, the car takes time t s.
Also, let's assume while covering this distance the car accelerated for t0 s.
From 0 s to t0 s, we know:
v=v0 m/s
u=0 m/s
a=9 m/s2
t=t0 s
By using v=u+at we get,
⇒v0=0+9t0=9t0
From t0 s to t s,
We know,
v=0 m/s
u=v0 m/s
a=−6 m/s2
t=(t−t0) s
By using v=u+at we get,
⇒0=v0−6(t−t0)
⇒v0=6(t−t0)
But v0=9t0
⇒9t0=6(t−t0)
⇒(t−t0)=t0×32
⇒t=t0×52
Applying s=ut+12at2 from 0 s to t0 s,
s1=0+12×9×(t0)2
s1=92t20
Applying s=ut+12at2 from t0 s to t s,
s2=(9t0)×(t−t0)+12×(−6)×(t−t0)2
s2=(9t0)×32t0−3(32t0)2
s2=32×92t20
We know,
s1+s2=1000 m=(1+32)×92t20
t20=445×1000
t0=9.42809 s
t=52×t0=23.57022 s