Kinematics Question 376
Question: A particle is moving along a straight line path according to the relation $ s^{2}=at^{2}+2bt+c $ s represents the distance travelled in t seconds and a, b, c are constants. Then the acceleration of the particle varies as
Options:
A) $ {{s}^{-3}} $
B) $ {{s}^{3/2}} $
C) $ {{s}^{-2/3}} $
D) $ s^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ s^{2}=at^{2}+2bt+c\text{ }\therefore \text{ 2s}\frac{ds}{dt}=2at+2b $
$ \text{or }\frac{ds}{dt}=\frac{at+b}{s},\text{ again differrentiating} $
$ \frac{d^{2}s}{dt^{2}}=\frac{as^{2}-{{( at+b )}^{2}}}{s^{3}}\text{ }\therefore \text{a=}\frac{d^{2}s}{dt^{2}}\propto {{s}^{-3}}. $
BETA
