Question

An athlete starts from rest with an acceleration $2m{s}^{-2}$ for sometime and then decelerates with $3m{s}^{-2}$ and finally comes to rest. If total length of track is 100m, then maximum speed attained by athlete is

• A. $4\sqrt{15}m{s}^{-1}$
• B. $2\sqrt{15}m{s}^{-1}$
• C. $10\sqrt{10}m{s}^{-1}$
• D. $20m{s}^{-1}$

Answer 1

The correct option is A $4\sqrt{15}m{s}^{-1}$

Given,

$u=0m/s$

$a=2m/s^2$

Lets consider, $v=$ maximum speed attained by a athlete

From 3rd equation of motion,

$2as=v^2-u^2$

$2\times 2\times s^{\prime }=v^2-0^2$

$s^{\prime }=\frac{v^2}{4}$. . . . . . . . . . .(1)

Now, athlete reach maximum speed and then decelerate with $3m/s^2$ and finally comes to rest,

From 3rd equation of motion,

$2as=v^2-u^2$

$-2\times 3\times s^{\prime \prime }=0^2-v^2$

$s^{\prime \prime }=\frac{v^2}{6}$. . . . . . . . . .(2)

According to question,

$s^{\prime }+s^{\prime \prime }=100$

$\frac{v^2}{4}+\frac{v^2}{6}=100$

$\frac{10v^2}{24}=100$

$v^2=240$

$v=\sqrt{240}m/s$

$v=4\sqrt{15}m/s$

The correct option is A.