Question

The position vector of a particle related to time t is given by \(\vec{r}=(10t\hat{i}+15t^2\hat{j}+7\hat{k})m\)
The direction of net force experienced by the particle is :

• A. Positive x-axis
• B. Positive y-axis
• C. Positive z-axis
• D. In x-y plane

Answer 1

The Correct Option is B

To determine the direction of the net force experienced by the particle, we need to understand its motion described by the position vector as a function of time: \(\vec{r}=(10t\hat{i}+15t^2\hat{j}+7\hat{k}) \, \text{m}\) .

1. First, we find the velocity vector \(\vec{v}\) , which is the first derivative of the position vector with respect to time (t) :
   \(\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}\left(10t\hat{i} + 15t^2\hat{j} + 7\hat{k}\right)\). \\ = (10\hat{i} + 30t\hat{j} + 0\hat{k}) \, \text{m/s}.
2. Next, we find the acceleration vector \(\vec{a}\) , which is the first derivative of the velocity vector with respect to time:
   \(\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}\left(10\hat{i} + 30t\hat{j}\right)\). \\ = (0\hat{i} + 30\hat{j} + 0\hat{k}) \, \text{m/s}^2.
3. The acceleration vector indicates that the particle experiences an acceleration only in the y-direction, since: \(\vec{a} = 30\hat{j} \, \text{m/s}^2\) .
4. According to Newton's second law, the net force \(\vec{F}\) on the particle is given by:
   \(\vec{F} = m\vec{a} \). Since the acceleration is directed along the positive y-axis, the force is also directed along the positive y-axis.

Thus, the direction of the net force experienced by the particle is along the positive y-axis.