Question

A particle moves along a straight line OX. At a time t (in seconds) the distance x (in meters) of the particle from 0 is given by:- $ x = 40 + 12t - (t^3) $ . How long would the particles travel before coming to rest?
A.16m
B.24m
C.40m
D.56m

Answer 1

Hint: Equations of motion relate the displacement of an object with its velocity, acceleration and time. $ s = v_t $ where s is the displacement, v the (constant) speed and t the time over which the motion occurred. $ s = s_0 + v_0 + \dfrac{1}{2}\times (at^2) $ a constant term $ (s_0) $ , followed by a first order term $ v_0(t) $ , followed by a second order term $ \dfrac{1}{2}\times (at^2) $ . Since the highest order is 2.

Complete step by step solution:
At t=0, particle is at, let's say x distance, from O;
Then putting t=0 in the given displacement-time equation we get;
 $
x=40+12t−t \\
x=40+12(0) − (0) \\
 =40\;m \:
 $
Particle comes to rest that means velocity of particle becomes zero after travelling certain displacement; let's say the time bet.
Then after differentiating the given displacement − time equation writ. Time we get velocity − time equation
 $
\Rightarrow v = u + at\\
v=12−3t \:
$
At time t=t (the time when the particle comes to rest):
 $
v=0 \\
\Rightarrow 12−3t =0 \\
\Rightarrow t=2s \;
$
Then, at t=2s we are at, let's say x
Distance from O;
Put this value of t (=2) in given displacement-time equation,
We get;
 $
\Rightarrow x=40+12t−t \\
=40+12(2) − (2) \\
=56\;m \:
 $
Further;
We have seen that the particle started his journey when it is at 40 m from the point O. And came to rest at 56 m from the point O.
Then the particle traveled a distance of:
 $ 56−40=16\;m $ .
Hence, Option (A) is the correct answer.

Note: $ s= ut + {\dfrac{1}{2}\times(at^2)} $
It is the expression that gives the distance covered by the object moving with uniform acceleration. The position of a particle moving along a straight line is given by
$ x(t)=BA[1−e^{At}] $ , where B is constant and A>0.