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

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.