Question

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

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