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

The correct option is A $16m$

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$;$

$x=40+12\left(0\right)-(0{)}^3=40m$

Particle comes to rest that means velocity of particle becomes zero after travelling certain displacement $;$ let's say the time be $t.$

then after differentiating the given displacement$-$time equation $wrt.$ time we get velocity$-$time equation

$v=12-3t^2$

at time $t=t$ $($the time when the particle comes to rest $):$

$v=0;$

$=>12-3t^2=0;$

$=>t=2s$

Then $,$at $t=2s$ we are at $,$ let's say $x^{\prime }$ distance from $O;$

put this value of$t(=2)$ in given displacement-time equation ,

we get$;$

$x^{\prime }=40+12\left(2\right)-(2{)}^3;$

$=56m$

Further$;$

We have seen that the particle started his journey when it is at $40m$ from the point $O.$

And came to rest at $56m$ from the point $O.$

then the particle traveled a distance of$:$

$56-40=16m.$

Hence,

option $\left(A\right)$ is correct answer.