Applications Of Derivatives Question 58
Question: The distance in seconds, described by a particle in t seconds is given by $ s=ae^{t}+\frac{b}{e^{t}} $ . Then acceleration of the particle at time t is
Options:
A) Proportional to t
B) Proportional to s
C) s
D) Constant
Show Answer
Answer:
Correct Answer: C
Solution:
Given that $ s=ae^{t}+\frac{b}{e^{t}} $ Differentiating w.r.t. time t, we get $ \frac{ds}{dt}= $ velocity $ =ae^{t}-\frac{b}{e^{t}} $ Again $ \frac{d^{2}s}{dt^{2}}= $ acceleration $ =ae^{t}+\frac{b}{e^{t}}=s $ .
BETA
