Q. If a particle is moving with velocity $v(t) = \cos{\pi t}$ along a straight line such that at $t = 0,\ s = 4$ its position function is given by
A) $\frac{1}{\pi}\cos{\pi t + 2}$
B) $- \frac{1}{\pi}\sin{\pi t + 4}$
C) $\frac{1}{\pi}\sin{\pi t + 4}$
D) None of these
Explanation
(c)
$\because v(t) = \cos{\pi t},$
$\Rightarrow \frac{ds}{dt} = \cos{\pi t}$
$\Rightarrow s = \frac{\sin{\pi t}}{\pi} + c,\ at\ t = 0,\ s = 4,\ c = 4$
$\Rightarrow s = \frac{\sin{\pi t}}{\pi} + 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.