Question

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

• A.
• B.
• C.
• D. None of above

Answer 1

The correct option is A

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 $u=\frac{7}{6}\frac{m}{s}$ and $a=\frac{1}{3}\frac{m}{s^2}$
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 2 sec = $s_{10}-s_8$
$=28.3-20=8.3m$