Question

Let position of a particle be expressed as $\vec r{\text{ }} = {\text{ }}({t^2}{\text{ }}\hat i{\text{ }} + {\text{ }}2t{\text{ }}\hat j){\text{ }}m$, (where $t$ is time in second). Its initial speed is
A. $2{\text{ }}m/s\:$
B. $ - 2{\text{ }}m/s\:$
C. $4{\text{ }}m/s\:$
D. $Zero\:$

Answer 1

Hint:We will initially differentiate the given expression of the position vector of the particle and by our basic concepts, it will give us the expression for the velocity vector. Then we will use the initial criteria and find the initial velocity vector. Then we will find the magnitude of the initial velocity vector which will give us the initial speed.

Formula used:
$|\vec a|{\text{ }} = {\text{ }}\sqrt {{a_x}^2{\text{ }} + {\text{ }}{a_y}^2{\text{ }} + {\text{ }}{a_z}^2} $

Complete step by step answer:
Here, the given position vector is
$\vec r{\text{ }} = {\text{ }}({t^2}{\text{ }}\hat i{\text{ }} + {\text{ }}2t{\text{ }}\hat j){\text{ }}m$
Differentiating both sides with respect to time, we get
$\dfrac{{d\vec r}}{{dt}}{\text{ }} = \dfrac{d}{{dt}}({t^2})\hat i{\text{ }} + \dfrac{d}{{dt}}(2t)\hat j$
After further evaluation, we get
$\dfrac{{d\vec r}}{{dt}}= 2t \hat i + 2 \hat j$
Now, we know $\dfrac{{d\vec r}}{{dt}}$ is nothing but the velocity vector $\vec v$,
Thus,
$\vec v = 2t \hat i + 2 \hat j$
Now, the criteria for initial point is just $t{\text{ }} = {\text{ }}0$
Thus, putting in this value, we get
$\vec v= 2 \hat j$
Thus,
Initial speed is,
$|\vec v|{\text{ }} = {\text{ }}\sqrt {{{(2)}^2}} $
After further evaluation, we get
$\therefore |\vec v|{\text{ }} = {\text{ }}2{\text{ }}m/s$

Hence, the correct answer is A.

Additional Information:
If we further another step; then, We will get the acceleration vector.Now, we see that the velocity is a constant vector.Which means that if we differentiate further.Then, we will arrive at a null vector.Which means that the particle moves at a constant speed and we can say that if the speed is constant, then there will be no change in speed and thus the acceleration is zero. This is also evident from our calculations as the magnitude of a null vector is also zero.

Note:Students often make mistakes in differentiating the vector. They also commit errors in misinterpreting the magnitude of a vector as just the absolute of the vector.Whereas the actual magnitude is driven by the formula we used above.