Q. A bullet emerge from a barrel of length 1.2 m with a speed of $640\ {ms}^{- 1}.$ Assuming constant acceleration, the approximate time that is spends in the barrel after the gun is fired is

A) $4\ ms$
B) $40\ ms$
C) $400\ \mu s$
D) $1\ s$
Explanation

(a) Given, $s = 1.2\ m,\ v = 640\ {ms}^{- 1},\ \ a = ?,\ \ u = 0;t = ?$
We have the third equation of motion
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2as = v^{2} - u^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ 2a \times 1.2 = 640 \times 640$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a = \frac{8\ \times \ 64\ \times \ 10^{3}}{3}$
And by first equation of motion
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = u + at$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t = \frac{v}{a} = \frac{15}{4} \times 10^{- 3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3.75 \times 10^{- 3}s\ \approx 4\ ms$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.