Question

A particle starts with a velocity of $2{\text{ms}}^{-1}$ and moves in a straight line with a retardation of $0.1{\text{ms}}^{-2}$. The first time at which the particle is $15\text{m}$ from the starting point is

• A. $10\text{s}$
• B. $20\text{s}$
• C. $30\text{s}$
• D. $40\text{s}$

Answer 1

The correct option is A $10\text{s}$

Taking the direction of motion to be positive direction
$u=2m/s$
$a=-0.1m/s^2(\because$ motion is retarded)
By using second equation of motion, we get
$15=2t-\frac{1}{2}\times \left(0.1\right)t^2$
$\left(0.1\right)t^2-4t+30=0$
On solving, we get $t=10\text{s}$ and $t=30\text{s}$
$\therefore$ For the first time particle will cover $15\text{m}$ in $10\text{s}$.