Applications Of Derivatives Question 246
Question: A particle is moving on a straight line, where its position s (in metre) is a function of time t (in seconds) given by $ s=at^{2}+bt+6,t\ge 0 $ . If it is known that the particle comes to rest after 4 seconds at a distance of 16 metre from the starting position $ (t=0) $ , then the retardation in its motion is
[MP PET 1993]
Options:
A) $ -1m/{{\sec }^{2}} $
B) $ \frac{5}{4}m/{{\sec }^{2}} $
C) $ -\frac{1}{2}m/{{\sec }^{2}} $
D) $ -\frac{5}{4}m/{{\sec }^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
Given equation $ s=at^{2}+bt+6 $ –(i)
Differentiating w.r.t. time, we get Velocity (v) $ =2at+b $ –(ii)
After 4sec, $ v=0 $ and distance $ s=16,\text{m} $
$ \therefore 0=2a\times 4+b\Rightarrow 8a+b=0 $ …………..(iii) and $ 16=16a+4b+6 $
therefore $ 16=16a+4(-8a)+6 $
$ \therefore $ $ a=-\frac{5}{8} $ But retardation in its motion is, $ 2a=\frac{5}{4}m/{{\sec }^{2}} $
$ \therefore $ Retardation $ =\frac{5}{4}m/s^{2} $ (Retardation itself means -ve).
BETA
