Question

A particle travels uniform velocity of 5m/s due north 3sec it stops in negligible time and then travels with uniform velocity of 10m/s due south for 7sec the avearage velocity of particle is

Answer 1

Answer:

If initial velocity be u and acceleration be a then distance in first 5 second will be according to S=ut+at

2

/2 i.e 10=5u+25a/2...(1)

And distance in 8 second will be S=8u+64a/2=8u+32a

So distance between t=5 and t=8 will be S−10=8u+32a−10

Which is given as 10m so 10=8u+32a−10 or 20=8u+32a.....(2)

Solving equation 1 and equation 2 we get a=1/3 and u=35/30

Now distance in 10 second is S

1

=10u+100a/2=10u+50a

So distance between t=8s and t=10s i.e. distance in last two seconds will be S

1

−S=10u+50a−(8u+32a)=2u+18a=

30

70

+

3

18

=

3

25

meter which is nearly 8.3meter .