Applications Of Derivatives Question 145
If the law of motion in a straight line is $ s=\frac{1}{2}a,t^2, $ then acceleration is
[MP PET 1991]
Options:
A) Constant
B) Proportional to t^2
C) Proportional to v
D) Proportional to s²
Show Answer
Answer:
Correct Answer: A
Solution:
$ s=\frac{1}{2}at^2 $
therefore $ 2s=2vt $
therefore $ 2\frac{ds}{dt}=v+t\frac{dv}{dt} $
therefore $ 2\frac{d^{2}s}{dt^{2}}=\frac{dv}{dt}+t.\frac{d^{2}v}{dt^{2}}+\frac{dv}{dt} $ But $ \frac{dv}{dt} $ = acceleration
therefore $ 2a=a+t\frac{da}{dt}+a $
therefore $ \frac{da}{dt}=0 $ or $ xy=4\times 4=16 $
But for the whole notation, $ t=0 $ is impossible, so that $ \frac{da}{dt}=0 $, i.e., a is constant.
BETA
