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

Answer: (A)

Step 1: Given that:

The initial velocity(u) of the body= $2m{s}^{-1}$

Retardation(-a) in the particle= $0.1m{s}^{-2}$

Displacement in the particle(s)= $15m$

Step 2: Calculation of time for the given displacement:

According to second equation of motion;

$s=ut+\frac{1}{2}a{t}^2$

Thus,

$15m=2m{s}^{-1}\times t-\frac{1}{2}\times 0.1m{s}^{-2}\times {\left(t\right)}^2$

$15=2t-\frac{t^2}{20}$

$15=\frac{40t-t^2}{20}$

$300=40t-t^2$

$t^2-40t+300=0$

$t^2-(30+10)t+300=0$

$(t^2-30t)-(10t-300)=0$

$t(t-30)-10(t-30)=0$

$Thus,$

$t=10secort=30sec$

Two times shows that,

In the first 10 sec, the particle will cover a displacement of $15m$ from the starting point and after the velocity becomes zero it moves in the backward direction. Thus, in the next $30sec$, the particle will again have a displacement of $15m$.

Thus,

In $10sec$ the particle will have a displacement of 15m firstly.

Hence,

Option C) 10sec is the correct option.