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 :
The direction of net force experienced by the particle is :
- Positive x-axis
- Positive y-axis
- Positive z-axis
- In x-y plane
The Correct Option is B
Solution and Explanation
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.
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