Question

Gogi was driving car at 20 m/s . He applied brakes which generate force equal to -500N. Find the distance after which the car stops . Its mass is 100 kg.​

Answer 1

Answer:

Given,

Speed=20m/s,d=50m,retardation=5m/s2

Let the ppossible reaction time be ts So the car moves 20tm And applied breaks with retardation 5m/s2

So, the total distance under retardation is ≤50m

Thus the distance under  the application of retardation =2×5202−02=s′⇒s′=40m

Therefore,  20ts+40≤=50⇒20ts≤50−40⇒20ts=10⇒ts≤21≤0.5s

Answer 2

Answer:

here

u=20m/s

V=0m/s

Mass=100kg

Force =-500N

Distance(s)=?

F=ma

-500N=100kg×a

-500/100=a

a=-5m/s², retardation=5m/s²

From 3rd equation of motion,

v²=u²+2as

(0m/s)²=(20m/s)²+2×(5m/s²)×s

0=400m²/s²+(-10m/s²×s

(-400m²/s²)/(-10m/s²)=s

=40m=s

hence distance =40 metres