Question

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

• A. $5 m/s^2$
• B. $-4 \, ms^2$
• C. $ -8 \, ms^2$
• D. 0

Answer 1

The Correct Option is B

To find the acceleration of the particle at \( t = 2 \, \text{s} \), we need to determine the particle's velocity and acceleration from the equations of motion provided.

The position coordinates of the particle as a function of time \( t \) are given by:

• \( x = 5t - 2t^2 \)
• \( y = 10t \)

To find the velocity components, we differentiate the position equations with respect to time \( t \).

Velocity in the x-direction:

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

Velocity in the y-direction:

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

Next, we find the acceleration components by differentiating the velocity equations with respect to time \( t \).

Acceleration in the x-direction:

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

Acceleration in the y-direction:

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

The total acceleration at any time is the vector sum of the x and y components of acceleration. At \( t = 2 \, \text{s} \), we have:

a = \sqrt{a_x^2 + a_y^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4 \, \text{m/s}^2

However, since \( a_y = 0 \), the acceleration is purely in the x-direction and negative.

Thus, the acceleration at \( t = 2 \, \text{s} \) is -4 \, \text{m/s}^2.

The correct answer is:

-4 \, \text{m/s}^2