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

Answer 1

The correct option is A $8.3m$

Step: 1 Find the relation of initial velocity and acceleration of the particle in first $5$ sec.
Formula Used: $S=ut+\frac{1}{2}a{t}^2$
Given that particle travels $10m$ in first $5$ sec
Let initial $(t=0)$ velocity of particle $u$
For first $5$ sec motion $s_5=10$ metre
$s_5=ut+\frac{1}{2}a{t}^2$
$10=5u+\frac{1}{2}a(5{)}^2$
$2u+5a=4\dots \dots \left(i\right)$

Step: 2 Find the relation of initial velocity and acceleration of the particle in first $8$ sec.
Formula Used: $S=ut+\frac{1}{2}a{t}^2$
For first $8$ sec of motion, $s_8=20$ meter.
$20=8u+\frac{1}{2}a(8{)}^2$
$2u+8a=5\dots \dots \left(ii\right))$

From equation (𝑖) and (𝑖𝑖), we get
$u=\frac{7}{6}m/s$ and $a=\frac{1}{3}m/s^2$

Step: 3 Find the distance travelled by the particle in $10$ sec.
Now distance travelled by particle in total $10$ sec
$s_{10}=u\times 10+\frac{1}{2}a(10{)}^2$
$s_{10}=\frac{7}{6}\times 10+\frac{1}{2}\left(\frac{1}{3}\right)(10{)}^2$ $s=28.3m$

Step: 4 Find the distance travelled by the particle in last $2$ sec.
The distance in last $2$ sec
$=s_{10}-s_8$
$=28.3-20$
$=8.3m$