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

Understanding the Problem

We are given the position vector of a particle:

\( r = 10t\hat{i} + 15t^2\hat{j} + 7t\hat{k} \)

We need to find the direction of the net force acting on the particle.

Solution

1. Velocity Vector (v):

The velocity vector is the first derivative of the position vector with respect to time:

\( v = \frac{dr}{dt} = \frac{d}{dt}(10t\hat{i} + 15t^2\hat{j} + 7t\hat{k}) \)

\( v = 10\hat{i} + 30t\hat{j} + 7\hat{k} \)

2. Acceleration Vector (a):

The acceleration vector is the first derivative of the velocity vector with respect to time (or the second derivative of the position vector):

\( a = \frac{dv}{dt} = \frac{d}{dt}(10\hat{i} + 30t\hat{j} + 7\hat{k}) \)

\( a = 30\hat{j} \)

3. Net Force (F):

According to Newton's second law, the net force is given by:

\( F = ma \)

Since \( a = 30\hat{j} \), we have:

\( F = m(30\hat{j}) = 30m\hat{j} \)

4. Direction of the Force:

The acceleration vector \( a = 30\hat{j} \) indicates that the acceleration is directed along the positive y-axis.

Since \( F = ma \), the force vector \( F = 30m\hat{j} \) is also directed along the positive y-axis.

Final Answer

The net force acting on the particle is directed along the positive y-axis.