Question

Starting from rest a particle moves in a straight line with acceleration $a=\left(25-t^2\right)^{1 / 2} m / s ^2$ for $0 \leq t \leq 5 s , a=\frac{3 \pi}{8} m / s ^2$ for $t >5\, s$. The velocity of particle at $t=7\, s$ is:

• A. $11 \, m / s$
• B. $22 \, m / s$
• C. $33\, m / s$
• D. $44 \, m / s$

Answer 1

Answer: (B)

$V _7= V _5+\frac{3 \pi}{8} \times 2$
To calculate $V _5$ i.e., velocity at end of $5 sec$.
$a=\left(25-t^2\right)^{1 / 2} \Rightarrow \frac{d v}{d t}=\left(25-t^2\right)^{1 / 2}$
$\int d v=\int 5\left(25-25 \sin ^2 \theta\right)^{1 / 2} \cos \theta d \theta\begin{cases}{l}t=5 \sin \theta \\ \frac{d t}{d \theta}=5 \cos \theta \\ t=0, \theta=0 \\ t=5, \theta=\frac{\pi}{2}\end{cases}$
$=\int\limits_0^{\pi / 2} 5 \times 5 \cos ^2 \theta d \theta=\frac{25 \pi}{4}$
$V_7=\frac{25 \pi}{4}+\frac{3 \pi}{4}=\frac{28 \pi}{4}=7 \pi=22 m / s$