Question

A MOTOR CYCLE RUNNING DUE WEST AT A VELOCITY OF 72KM/H IS BROUGHT TO REST IN 2 SEC BY APPLICATION OF BRAKES.DISTANCE COVERD BY THE MOTOR CYCLE IN THESE 2 SEC IS

1. 40 M
2. 20 M
3. 15 M
4. 10 M

Answer 1

u=72kmph=20m/s

v=0m/s

t=2s

a=v-u/t                                                          OPTION-2

   = -10m/s²

S=ut+1/2at²

     =20×2+1/2×(-10)×4

     =40-20

     S=20m

  

Answer 2

Initial velocity(u) = 72KM/H = 20 m/sec.
Final velocity(v)= 0 m/sec
Time taken (t) = 2 sec.
Acceleration(a) = ? 
Displacement(s) = ?
From v = u+at 
   0 = 20 + a*2
 on solving ,we get  
 a = -10 m/sec^2
From s = ut+$\frac{at^2}{2}$
  s = 20*2 - $\frac{10*2^2}{2}$
s = 20m.