Question

A particle covers $10m$ in first $5s$ and $10m$ in next $3s$. Assuming constant acceleration. Find initial speed, acceleration and distance covered in next $2s$.

Answer 1

It is given that a particle covers 10m in first 5s and 10m in next 3s. so using the equation of motion

Case I

$s=ut+\frac{1}{2}a{t}^2$

$10=5u+\frac{1}{2}a{\left(5\right)}^2$

$20=10u+25a$

$4=2u+5a..............\left(1\right)$

Case 2

In next 3s the particle covers more 10m distance. So

$20=8u+\frac{1}{2}a{\left(8\right)}^2$

$5=2u+8a.........\left(2\right)$

On solving equation (1) and (2)

$4=2u+5a$

$5=2u+8a$

$a=\frac{1}{3}m/s^2$

Put the value of a in equation (1)

$u=\frac{7}{6}m/s$

Now to find distance in next 10 s. total time will be 10s

$s=\frac{7}{6}\times 10+\frac{1}{2}\times \frac{1}{3}\times {\left(10\right)}^2$

$s=28.33m$

Distance travelled in next 2 sec

$s=28.33-20=8.33m$