Question

A particle starts from rest, moves with constant acceleration for $15s$. If it covers $s_1$ distance in first $5s$ their distance $s_2$ in next $10s$, then find the relation between $s_1$ and $s_2$.

Answer 1

if acceleration be $a$ then using equation $S=ut+a{t}^2/2$ and putting initial velocity $u=0$ we get $s_1=a{5}^2/2=25a/2$..................... (1)

put $t=15$ we get distance in $15second$ as $S=a{15}^2/2=225a/2$

Now distance travelled in $last$ $10seconds$ will be $\text{distance in 15 second - distance in 5 second}=S-s_1=225a/2-25a/2=200a/2=100a$

which is nothing but $8\times 25a/2$ or $8s_1$

So $s_2=8s_1$