Question

At t=0, a particle at (1,0,0) moves towards point (4,4,12) with a constant velocity of magnitude 65m/s. The position of the particle is measured in metres and time in sec. Assuming constant velocity, the position of the particle at t=2sec.
$\begin{aligned} & \text{A}\text{.}\left( 13\hat{i}-120\hat{j}+40\hat{k} \right)m. \\\ & \text{B}.\left( 40\hat{i}+31\hat{j}-120\hat{k} \right)m. \\\ & \text{C}.\left( 13\hat{i}-40\hat{j}+12\hat{k} \right)m. \\\ & \text{D}.\left( 31\hat{i}+40\hat{j}+120\hat{k} \right)m. \\\ \end{aligned}$

Answer 1

Firstly we will find the unit vector in the direction of the particle from ( 1,0,0) to (4,4,12). A body is said to be moving with uniform acceleration if velocity vectors suffer the same change in the same interval of time. We will use the formula for the final position of the particle after time t in vector form and will obtain the answer.
Formula used:
${{\vec{r}}_{f}}={{\vec{r}}_{i}}+\vec{v}\times 2$

Complete answer:
Firstly we will find the unit vector in direction of ( 1,0,0) to (4,4,12)
$\begin{aligned} & =\dfrac{\left( 4-1 \right)\hat{i}+\left( 4-0 \right)\hat{j}+\left( 12-0 \right)\hat{k}}{13} \\\ & =3\hat{i}+4\hat{j}+12\hat{k} \\\ \end{aligned}$
Now we will find the velocity
$\begin{aligned} & \vec{v}=65\times \dfrac{\left( 3\hat{i}+4\hat{j}+12\hat{k} \right)}{13} \\\ & =15\hat{i}+20\hat{j}+60\hat{k} \\\ \end{aligned}$
Hence position of the particle at t=2 sec
$\begin{aligned} & {{{\vec{r}}}_{f}}={{{\vec{r}}}_{i}}+\vec{v}\times 2 \\\ & =31\hat{i}+40\hat{j}+120\hat{k} \\\ \end{aligned}$

So, the correct answer is “Option D”.

Additional Information:
Position vector: a vector which gives the position of a particle with respect to origin of a co-ordinate system is known as position vector.
Displacement vector: It is the vector in which an object has changed its direction. It also tells how much object has been moved in a given time interval.
A body is said to be moving with uniform acceleration if velocity vectors suffer the same change in the same interval of time.
Each rectangular component of velocity of an object moving with uniform acceleration in a plane depends upon time as if it were the velocity vector of one –dimensional uniformly accelerated motion.
A uniform motion in two dimensions can be expressed as the sum of two uniform motions along two mutually perpendicular directions.
If the motion suffers equal displacements in an equal interval of time it is said to be with uniform velocity.

Note:
A unit vector in the direction of a given vector is found by dividing the given vector by its magnitude or its modulus.
$\vec{A}=\dfrac{{\vec{A}}}{\left| A \right|}$
In uniform acceleration, the position vector at time t.
The motion in a plane with uniform acceleration is treated as the superposition of two separate simultaneous one-dimensional motions along two perpendicular directions.