Kinematics Question 364
Question: A particle moves along a straight line OX. At a time t (in second) the distance x (in metre) of the particle from O it’s given by $ x=40+12t-t^{3} $ . How long would the particle travel before coming to rest?
Options:
A) 24 m
B) 40 m
C) 56 m
D) 16 m
Show Answer
Answer:
Correct Answer: C
Solution:
When particle comes to rest, $ \text{V=0=}\frac{dx}{dt}\text{=}\frac{d}{dt}( 40+12t-{{t}^{3}} ) $
$ \Rightarrow 12-3{{t}^{2}}=0 $
$ \Rightarrow {{t}^{2}}=\frac{12}{3}=4\text{ }\therefore \text{t=2 sec} $
therefore distance travelled by particle before coming to rest, $ x=40+12t-t^{3}=40+12\times 2-{{( 2 )}^{3}}=56m $
