Question

At time t = 0 a particle starts travelling from a height \(7\hat z cm\) in a plane keeping \(z\) coordinate constant. At any instant time it's position along the \(x\) and \(y\) directions are defined as \(3t\) and \(5t^3\) respectively. At \(t = 1s\) acceleration of the particle will be

• A. \(-30 \hat {y}\)
• B. \(30\hat {y}\)
• C. \(3 \hat {x}+ 15\hat {y}\)
• D. \(3\hat{x}^+15\hat{y}+7\hat{z}\)

Answer 1

The Correct Option is B

To find the acceleration of the particle at time \( t = 1 \) second, we need to first understand the equations that describe the motion of the particle and calculate the derivatives accordingly.

The position of the particle in the \( x \) direction is given by the equation x(t) = 3t.

The position of the particle in the \( y \) direction is y(t) = 5t^3.

The \( z \) coordinate remains constant at \( 7 \, \text{cm} \), and thus does not affect acceleration.

Step 1: Compute the velocity components.

• Velocity in the \( x \)-direction: \( v_x = \frac{dx}{dt} = \frac{d}{dt}(3t) = 3 \)
• Velocity in the \( y \)-direction: \( v_y = \frac{dy}{dt} = \frac{d}{dt}(5t^3) = 15t^2 \)

At \( t = 1s \), the velocity components are:

• v_x = 3\
• v_y = 15 \times (1)^2 = 15\

Step 2: Compute the acceleration components.

• Acceleration in the \( x \)-direction: \( a_x = \frac{dv_x}{dt} = \frac{d}{dt}(3) = 0 \)
• Acceleration in the \( y \)-direction: \( a_y = \frac{dv_y}{dt} = \frac{d}{dt}(15t^2) = 30t \)

At \( t = 1s \), we find:

• a_x = 0\
• a_y = 30 \times (1) = 30\

Thus, the acceleration of the particle at \( t = 1s \) is in the \( y \) direction: 30\hat{y}\.

Conclusion:

The acceleration of the particle at \( t = 1 \) second is 30\hat{y}\, which corresponds to the correct option.