Question

An automobile moving with a speed 60kmph can break to stop at a distance 20m. A car is moving twice as fast as this, the stopping distance will be?

Answer 1

Given:
u=60km/hr,
v=0,
s=20m=20/1000m=0.02km

According to kinematic equations of motions,
v²-u²=2as
0-(60)²=2(a)(0.02)
-3600/0.04=a
-90000km/hr²=a

speed becomes twice
Therefore,
speed,u=120km/hr
v=0
a=-90000km/hr²
s=?

According to kinematic equations of motions,
v²-u²=2as
0-(120)²=2(-90000)s
-14400/-180000=s
s=0.08km or s=80m

Hence, if the car is going twice as fast, then, the stopping distance will be 0.08km or 80m.