Question

The x and y coordinates of the particle at any time are x = 5t – 2t² and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2 s is

• A. \(0\)
• B. \(5 \ m/s^2\)
• C. \(-4 \ m/s^2\)
• D. \(-8 \ m/s^2\)

Answer 1

The Correct Option is C

To determine the acceleration of the particle at \( t = 2 \, \text{s} \), we need to find the second derivative of the position functions \( x(t) \) and \( y(t) \) with respect to time to get the components of acceleration.

The position functions provided are:

• \( x(t) = 5t - 2t^2 \)
• \( y(t) = 10t \)

Step 1: Differentiate to find velocity components

The velocity component in the x-direction, \( v_x(t) \), is given by the first derivative of \( x(t) \) with respect to time \( t \):

\[ v_x(t) = \frac{dx}{dt} = \frac{d}{dt}(5t - 2t^2) = 5 - 4t \]

The velocity component in the y-direction, \( v_y(t) \), is given by the first derivative of \( y(t) \) with respect to time \( t \):

\[ v_y(t) = \frac{dy}{dt} = \frac{d}{dt}(10t) = 10 \]

Step 2: Differentiate to find acceleration components

The acceleration component in the x-direction, \( a_x(t) \), is given by the derivative of \( v_x(t) \) with respect to time \( t \):

\[ a_x(t) = \frac{dv_x}{dt} = \frac{d}{dt}(5 - 4t) = -4 \]

The acceleration component in the y-direction, \( a_y(t) \), is given by the derivative of \( v_y(t) \) with respect to time \( t \):

\[ a_y(t) = \frac{dv_y}{dt} = \frac{d}{dt}(10) = 0 \]

Step 3: Calculate the total acceleration

At \( t = 2 \, \text{s} \), the acceleration components are:

• \( a_x = -4 \, \text{m/s}^2 \)
• \( a_y = 0 \, \text{m/s}^2 \)

Since acceleration in the y-direction is zero, the acceleration of the particle is purely in the x-direction. Therefore, total acceleration is:

\[ a = a_x = -4 \, \text{m/s}^2 \]

Conclusion: The acceleration of the particle at \( t = 2 \, \text{s} \) is -4 \, \text{m/s}^2. This matches with option: -4 \, \text{m/s}^2.