NEET Physics Motion in a Straight Line / सरल रेखा में गति NEET PYQ-One Dimensional Motion

The position x of a particle with respect to time t along x-axis is given by \[x=9{{t}^{2}}-{{t}^{3}}\] where x is in metre and t in second. What will be the position of this particle when it achieves maximum speed along the \[+x\] direction? [AIPMT (S) 2007]

A) 32 m

B) 54 m

C) 81 m

D) 24 m

Correct Answer: B

Solution :

Key Idea: At the instant when speed is maximum, its acceleration is zero.

Given, die position \[x\] of a particle with respect to time \[t\] along x-axis

\[x=9{{t}^{2}}-{{t}^{3}}\] ...(i)

Differentiating Eq. (i), with respect to time, we get speed, i.e.,

\[v=\frac{dx}{dt}=\frac{d}{dt}(9\,{{t}^{2}}-{{t}^{3}})\]

or \[v=18\,t-3{{t}^{2}}\] ...(ii)

Again differentiating Eq. (ii), with respect to time, we get acceleration, i.e.,

\[a=\frac{dv}{dt}=\frac{d}{dt}(18\,t-3{{t}^{2}})\]

or \[a=18-6t\] ...(iii)

Now, when speed of particle is maximum, its acceleration is zero, i.e.,

\[a=0\]

i.e., \[186t=0\] or \[t=3s\]

Putting in Eq. (i), we obtain position of particle at that time

\[x=9{{(3)}^{3}}-{{(3)}^{3}}=9(9)-27\]

\[=81-27=54\,m\]