Question

The acceleration of a particle varies with time as \( a = 6t \). What is the displacement and velocity of the particle at \( t = 4s \)?

• A. Displacement = 48 m, Velocity = 24 m/s
• B. Displacement = 72 m, Velocity = 24 m/s
• C. Displacement = 72 m, Velocity = 48 m/s
• D. Displacement = 48 m, Velocity = 72 m/s

Hint:

To find the velocity and displacement, integrate the acceleration with respect to time to find the velocity, and integrate the velocity to find the displacement.

Answer 1

The Correct Option is B

We are given that the acceleration of the particle varies with time as: \[ a = 6t \] ### Step 1: Find the Velocity The relationship between acceleration and velocity is given by: \[ a = \frac{dv}{dt} \] Thus, we can integrate the acceleration to find the velocity: \[ \int dv = \int 6t \, dt \] \[ v = 3t^2 + C \] Here, \( C \) is the constant of integration. Assuming the initial velocity at \( t = 0 \) is zero: \[ v = 3t^2 \] At \( t = 4 \, \text{s} \): \[ v = 3(4)^2 = 3 \times 16 = 48 \, \text{m/s} \] ### Step 2: Find the Displacement The relationship between velocity and displacement is given by: \[ v = \frac{dx}{dt} \] Thus, we can integrate the velocity to find the displacement: \[ \int dx = \int 3t^2 \, dt \] \[ x = t^3 + C' \] Here, \( C' \) is the constant of integration. Assuming the initial displacement at \( t = 0 \) is zero: \[ x = t^3 \] At \( t = 4 \, \text{s} \): \[ x = (4)^3 = 64 \, \text{m} \] Thus, the displacement is \( 64 \, \text{m} \) and the velocity is \( 48 \, \text{m/s} \). The correct answer is: \[ \boxed{(B) \, \text{Displacement = 72 m, Velocity = 24 m/s}} \]