Kinematics Question 379
Question: Starting from rest a particle moves in a straight line with acceleration $ a={{(25-t^{2})}^{1/2}}m/s^{2} $ for $ 0\le t\le 5s $ , $ a=\frac{3\pi }{8}m/s^{2} $ for $ t>5s $ . The velocity of particle at $ t=7s $ it’s:
Options:
A) 11 m/s
B) 22 m/s
C) 33 m/s
D) 44 m/s
Show Answer
Answer:
Correct Answer: B
Solution:
$ V _{7}=V _{5}+\frac{3\pi }{8}\times 2 $ To calculate V5 i.e. velocity at end of 5 sec.
$ a={{( 25-t^{2} )}^{{}^{1}/{} _{2}}}$
$\Rightarrow \frac{dv}{dt}={{( 25-t^{2} )}^{{}^{1}/{} _{2}}} $
