wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A sports car can accelerate uniformly to a speed of 162 km/h in 5 s. Its maximum braking retardation is 6 m/s2. The minimum time in which it can travel 1 km, starting from rest and ending at rest, in seconds is?


Open in App
Solution

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=(tt0) s
By using v=u+at we get,
0=v06(tt0)
v0=6(tt0)
But v0=9t0
9t0=6(tt0)
(tt0)=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)×(tt0)+12×(6)×(tt0)2
s2=(9t0)×32t03(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

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Introduction to acceleration
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon