Question

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

• A. $8.3\text{m}$
• B. $9.3\text{m}$
• C. $10.3\text{m}$
• D. $28.3\text{m}$

Answer 1

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

Let initial velocity of particle be $=u$
For first $5\text{sec}$ motion, $s_5=10\text{metre}$
As we know that
$s=ut+\frac{1}{2}a{t}^2\Rightarrow 10=5u+\frac{1}{2}a(5{)}^2$
$\Rightarrow 2u+5a=4$ ... (i)
For first $8\text{sec}$ of motion $s_8=20\text{metre}$
Similarly,
$20=8u+\frac{1}{2}a(8{)}^2\Rightarrow 2u+8a=5$ ... (ii)
By solving the above two equations, we get
$u=\frac{7}{6}\text{m/s}$ and $a=\frac{1}{3}{\text{m/s}}^2$
Now distance travelled by particle in total $10\text{sec}$ is given by
$s_{10}=u\times 10+\frac{1}{2}a(10{)}^2$
By substituting the value of $u$ and $a$ we will get
$s_{10}=28.3\text{m}$
So, the distance in last $2\text{sec}=s_{10}-s_8$
$=28.3-20=8.3\text{m}$