A particle travels 10m in the first 5 sec and 10m in the next 3 sec. Assuming constant acceleration what is the distance travelled in the next 2 sec?

s=ut + $\frac{1}{2}$at²

  = 5u +25a/2  .....(1)

Distance in 8 sec = 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 seconds is s₁​=10u+100a/2=10u+50a

So the distance between t=8s and t=10s i.e. distance in the last two seconds will be s₁−s

=10u+50a−(8u+32a)

=2u+18a=3070​+318​

=325​ metres which is nearly 8.3 metres