Question

A motorcyclist moving with a speed of 10 m/s, saw a puppy standing on the road 25 m ahead. He applies brake and manages to stop the motorcycle just before it reaches the puppy. What must be the deceleration of the motorcycle?

Answer 1

Answer:

then distance in t sec will be 10t meter

⇒ remaining distance = 10 meter - 10 t

using v2=u2−2as

we get s=2au2−v2=2a(10)2−0=a50

⇒a50=10−10t _______ (1)

Case II 

u=20 m/s & S0=30 meter

distance in t sec will be 20 t meter

distance during retardation S=2au2−v2=2a(20)2−0

S=a200

S0=30m⇒a200+20t=30

          a50+5t=7.5 _______ (2)

using (1) a50+10t=10

            -           -         -

         ___________________

           0-5t= -2.5 ⇒ subtracting t=0.5 second