Applications Of Derivatives Question 53
Question: If the distance -s- traveled by a particle in time t is $ s=a\sin t+b\cos 2t $ , then the acceleration at t = 0 is
Options:
A) a
B) - a
C) 4b
D) - 4b
Show Answer
Answer:
Correct Answer: D
Solution:
Given $ s=a\sin t+b\cos 2t $
$ \therefore $ $ \frac{ds}{dt}=a\cos t-2b\sin 2t $
$ \frac{d^{2}s}{dt^{2}}=-a\sin t-4b\cos 2t $ At $ t=0 $ , $ \frac{d^{2}s}{dt^{2}}=-a\sin 0{}^\circ -4b\cos 0{}^\circ =-4b $ .
BETA
