Question

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

• A. 8.3 m
• B. 9.3 m
• C. 10.3 m
• D. None of above

Answer 1

The correct option is A 8.3 m

let initial (t = 0) velocity of particle = u

For first 5 sec motion $s_5=10$ metre

$s=ut+\frac{1}{2}a{t}^2\Rightarrow 10=5u+\frac{1}{2}a(5{)}^2$

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

For first 8 sec of motion $s_8=20$ metre

$20=8u+\frac{1}{2}a(8{)}^2\Rightarrow 2u+8a=5.....(ii)$

By solving u = $\frac{7}{6}m/s$ and $a=\frac{1}{3}m/s^2$

Now distance travelled by particle in total 10 sec.

$s_{10}=u\times 10+\frac{1}{2}a(10{)}^2$

By substituting the value of u and a we will get $s_{10}=28.3m$

so the distance in last $2sec=s_{10}-s_8$

$=28.3-20=8.3m$