Question

13. A particle experiences constant acceleration for
20 second after starting from rest. If it travels a
distance S, in the first 10 seconds and distance S2 in
the next 10 seconds, then find the relation between S
Ans. S2 = 3S1
and S2.​

Answer 1

Answer:

S2=3S1

Step-by-step explanation:

Let  the  constant  acceleration  be  a

For  the  first 10 seconds ,

  t=10

  u=0

   v=u+at=0+a*10=10a

Therefore,

                  S1=ut+1/2$at^{2}$

                      =0+1/2*a*100

                     S1 =50a

For  the  next 10 seconds,

  t=10

  u=10a

Therefore,

                             S2=ut+1/2$at^{2}$

                                  =10a*10+1/2*a*100

                                  =100a+50a

                                  S2=150a

                                   S2=3*50a

                                    S2=3S1

  Hence we  get  the  relation  S2=3S1