Question

An
automobile
travelling
with
a
uniform
speed
of
60
km/h
on


applying
the
brakes
comes
to
a
stop
at
a
distance
of
20
m
with


uniform
retardation.
If
the
car
doubles
its
initial
speed,
what
will
be

the
new
stopping
distance?​

Answer 1

Answer:

Given,

Initial Velocity (u) = 60 km/hr

Final Velocity (v) = 0 km/hr

Distance Travelled (S) = 20 m

= 0.02 km

Retardation (a) = v² = u² - 2aS

= 0 = (60)² - [2*a*0.02]

= 0.04a = 3600

= a = 90000 km/hr²

Now,

Initial Velocity (u) = 120 km/hr

Final Velocity (v) = 0 km/hr

Retardation (a) = 90000 km/hr²

Here, we are assuming that the car retards at the same rate like the last time.

= v² = u² - 2aS

= 0 = (120)² - [2*90000*S]

= 180000S = 14400

= S = 0.008 km

or,

= S = 8 m

Answer: Assuming that the car retards at the same rate, it'll travel upto 0.008 km or 8 m.

I hope this helps.