Question

A particle travels $10\text{m}$ in first $5s$ and $10\text{m}$ in next $3s$. Assuming constant acceleration what is the displacement travelled in next $2sec$

• A. $8.3\text{m}$
• B. $9.3\text{m}$
• C. $10.3\text{m}$
• D. None of the above

Answer 1

The correct option is A $8.3\text{m}$

Let initial $(t=0)$ velocity of particle $=u$
For first $5s$ of motion $s_5=10\text{m}$ , so by using $s=ut+\frac{1}{2}a{t}^2$
$10=5u+\frac{1}{2}a(5{)}^2\Rightarrow 2u+5a=4\dots \dots (i)$
For first time $8s$ of motion $s_8=20\text{m}$
$20=8u+\frac{1}{2}a(8{)}^2\Rightarrow 2u+8a=5\dots \dots (ii)$
By solving $\left(i\right)$ and $\left(ii\right)$ $u=\frac{7}{6}\text{m/s},a=\frac{1}{3}{\text{m/s}}^2$
Now distance travelled by particle in total $10sec$. $s_{10}=u\times 10+\frac{1}{2}a(10{)}^2$
By substituting the value of $u$ and $a$ we will get $s_{10}=28.3\text{m}$
So the distance in last $2s$ $=s_{10}-s_8=28.3-20=8.3\text{m}$