Question

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

Answer 1

Speed is the rate at which an object moves. It’s a very basic concept in motion and all about how fast or slow an object can move. Speed is simply Distance divided by the time where Distance is directly proportional to Velocity when time is constant. Problems related to Speed, Distance, and Time, will ask you to calculate for one of three variables given.
In the given question the distance ‘$x$’ is a function of time ‘$t$’. It means with time ‘$t$’ distance is changing. We have to find the distance traveled by the particle to come to rest. It means we have to say that velocity is zero after travel.

Complete step by step solution:
The process of finding the derivatives is called differentiation. The inverse process is called anti-differentiation. Let, the derivative of a function be $y=f(x)$. It is the measure of the rate at which the value of $y$ changes with respect to the change of the variable $x$. It is known as the derivative of the function “$f$”, with respect to the variable $x$.
Velocity $(v)=\dfrac{d(x)}{d(t)}$
Where ‘x’ is the distance travelled and ‘$t$’ is the time taken to cover that distance.
This will give you the distance covered per unit time so that we can analyse any distance covered in any interval of time.

The given equation is
$x=40+12t-{{t}^{3}}$
Differentiate $x$ w.r.t $t$
$v=\dfrac{dx}{dt}=0+12-3{{t}^{2}}$
$v=12=3{{t}^{2}}$
At rest $v=0$
$0=12=3{{t}^{2}}\Rightarrow 3{{t}^{2}}=12$
$t=\pm 2$sec.
The value of t is positive
So distance, $x=40+12t-{{t}^{3}}$
$x=40+12(2)-{{(2)}^{3}}$
$x=40+24-8$
$x=56m$
We have seen that the particle started its journey when it is at $40m$ from the point $O$.
And came to rest at $56m$ from the point $O$.
then the particle travelled a distance of:
$56-40=16m$

Hence, Option (D) is the correct answer.

Note: Students have clear understanding about distance and displacement. In this question distance and displacement is the same due to straight line motion.
Students should know how to differentiate and when the body comes to rest means $v=0$.