Question

The position vector of a particle is determined by the expression $\vec r = 3{t^2}\hat i + 4{t^2}\hat j + 7\hat k$ .The distance traversed in first 10 sec is
A.500m
B.300m
C.150m
D.100m

Answer 1

Hint: Firstly, we will use the initial position at time = 0. Then we will put the value of time i.e, $t = 10$ . We will find the position at \[t = 0\] .Then we will find out the displacement vector by subtracting the final position vector at time \[t = 0\] from the initial position vector at time $t = 10$ . We will find only the magnitude of the displacement by taking mod on both sides.

Complete step by step solution
The position of an object is defined as its linear distance as well as its direction with respect to a preassigned reference point. Position is a physical quantity having both magnitude and direction. So, it is a vector quantity. It is called position vector and denoted by the symbol $\vec r$ .
Given that,
Position vector of particle is $\vec r = 3{t^2}\hat i + 4{t^2}\hat j + 7\hat k$
So, the position vector is the function of time.
Let, $\vec r(i) = $ initial position vector when time $(t) = 0$
 $\therefore \vec r(i) = \vec r(t = 0) = 7\hat k.....(i)$
 $\vec r(f) = $ final position vector at time \[\left( t \right) = 10\]
 $\therefore \vec r(f) = \vec r(t = 10) = 3 \times {10^2}\hat i + 4 \times {10^2}\hat j + 7\hat k....(ii)$
Displacement is defined as the change in position of a moving body in a fixed direction.
So we can write that,
Displacement vector $(D) = \vec r(f) - \vec r(i)$
 $
  \left| D \right| = \left| {300\hat i + 400\hat j + (7 - 7)\hat k} \right| \\
   \Rightarrow \left| D \right| = \sqrt {{{300}^2} + {{400}^2}} \\
 $
 $D = 500m$
Hence, distance traversed in first 10 sec is 500m (Option-A)

Note: One may think that after putting the value of $t = 10$ one can obtain the displacement vector. But this is actually a position vector. The displacement vector of a particle is the vector difference between its final and initial position vectors.
Additional Information:
For any particle in three-dimensional space, the displacement is represented by the straight line joining the initial and the final positions of the particle. If a particle travels from the point ${P_1}({x_1},{y_1},{z_1})$ to the point ${P_2}({x_2},{y_2},{z_2})$ then $\vec D$ represents its displacement. The magnitude of the displacement is given by $D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} $