Question

A particle moves in such a manner that $x = At$, $y = B{t^3} - 2t$ and $z = C{t^2} - 4t$ where $x$, $y$, $z$ are measured in metres and $t$ is measured in seconds, and $A$, $B$ and $C$ are unknown constants. Given that the velocity of the particle at $t = 2\,{\text{s}}$ is $\vec v = \left( {\dfrac{{d\vec r}}{{dt}}} \right) = 3\hat i + 22\hat j\,{\text{m/s}}$, determine the velocity of the particle at $t = 4\,{\text{s}}$.
A. $8\hat i + 94\hat j + 4\hat k\,{\text{m/s}}$
B. $6\hat i + 94\hat j + 6\hat k\,{\text{m/s}}$
C. $3\hat i + 94\hat j + 4\hat k\,{\text{m/s}}$
D. $3\hat i + 92\hat j + 4\hat k\,{\text{m/s}}$

Answer 1

Use the formula for velocity of a particle in terms of derivative of the position vector of the particle with respect to time. First determine the position vector of the particle at any time and then use this formula for velocity and determine velocity vector of the particle at any time t. Using this expression for velocity vector, determine the velocity of the particle at time 2 seconds and compare this expression with the given expression for velocity at 2 seconds. Determine the values of the constants and finally evaluate the expression for velocity of particle at time 4 seconds.

Formula used:
The velocity vector $\vec v$ is given by
$\vec v = \dfrac{{d\vec r}}{{dt}}$ …… (1)
Here, $d\vec r$ is the change in position vector in time $dt$.

Complete step by step answer:
We have given that the position vectors of a particle in X, Y and Z directions are
$x = At$
$\Rightarrow y = B{t^3} - 2t$
$\Rightarrow z = C{t^2} - 4t$
Here, $A$, $B$ and $C$ are unknown constants. The position vector of the same particle becomes
$\vec r = \left( {At} \right)\hat i + \left( {B{t^3} - 2t} \right)\hat j + \left( {C{t^2} - 4t} \right)\hat k$
Let us first determine the expression for the velocity vector of the particle at any time $t$. Substitute $\left( {At} \right)\hat i + \left( {B{t^3} - 2t} \right)\hat j + \left( {C{t^2} - 4t} \right)\hat k$ for $\vec r$ in equation (1).
$\vec v = \dfrac{{d\left[ {\left( {At} \right)\hat i + \left( {B{t^3} - 2t} \right)\hat j + \left( {C{t^2} - 4t} \right)\hat k} \right]}}{{dt}}$
$\Rightarrow \vec v = \left( A \right)\hat i + \left( {3B{t^2} - 2} \right)\hat j + \left( {2Ct - 4} \right)\hat k$ …… (2)
This is the expression for the velocity vector of the particle at any time.

Substitute $2\,{\text{s}}$ for $t$ in the above equation.
$\Rightarrow \vec v = \left[ A \right]\hat i + \left[ {3B{{\left( {2\,{\text{s}}} \right)}^2} - 2} \right]\hat j + \left[ {2C\left( {2\,{\text{s}}} \right) - 4} \right]\hat k$
$\Rightarrow \vec v = \left[ A \right]\hat i + \left[ {12B - 2} \right]\hat j + \left[ {4C - 4} \right]\hat k$ …… (3)
We have given that the velocity of the particle at time $t = 2\,{\text{s}}$ is
$\vec v = 3\hat i + 22\hat j\,{\text{m/s}}$
Comparing this equation for velocity at $t = 2\,{\text{s}}$ with the equation (3), we get
$A = 3$
And
$12B - 2 = 22$
$\Rightarrow B = 2$
And
$4C - 4 = 0$
$\Rightarrow C = 1$
Let us now calculate the velocity of the particle at time $t = 4\,{\text{s}}$.
Substitute $3$ for $A$, $2$ for $B$, $1$ for $C$ and $4\,{\text{s}}$ for $t$ in equation (2).
$\Rightarrow \vec v = \left[ 3 \right]\hat i + \left[ {3\left( 2 \right){{\left( {4\,{\text{s}}} \right)}^2} - 2} \right]\hat j + \left[ {2\left( 1 \right)\left( {4\,{\text{s}}} \right) - 4} \right]\hat k$
$\therefore \vec v = 3\hat i + 94\hat j + 4\hat k$
Therefore, the velocity of the particle at $t = 4\,{\text{s}}$ is $3\hat i + 94\hat j + 4\hat k$.

Hence, the correct option is C.

Note: The students may get confused while comparing the velocity vector at time 2 seconds because in the given expression for velocity at 2 seconds, the component of the unit vector in z direction is zero. At such time, we have to compare the unit vector of velocity with zero as there is no component of unit vector in z direction.