Question

A point initially at rest moves along x-axis. Its acceleration varies with time as \( a = (6t + 5)\, \text{m/s}^2 \). If it starts from origin, the distance covered in 2 s is:

• A. 20 m
• B. 18 m
• C. 16 m
• D. 25 m

Hint:

When acceleration is time-dependent, integrate step-by-step: \(a \to v \to x\), applying initial conditions at each stage.

Answer 1

The Correct Option is B

Concept: Acceleration is the derivative of velocity: \[ a = \frac{dv}{dt}, \quad v = \int a \, dt \] Position is obtained by integrating velocity.

Step 1: Find velocity.
\[ v = \int (6t + 5)\, dt = 3t^2 + 5t + C \] Since initially at rest \(v=0\) at \(t=0\): \[ C = 0 \] \[ v = 3t^2 + 5t \]

Step 2: Find displacement.
\[ x = \int v\, dt = \int (3t^2 + 5t)\, dt \] \[ x = t^3 + \frac{5}{2}t^2 + C \] At \(t=0\), \(x=0 \Rightarrow C=0\)

Step 3: Distance in 2 s.
\[ x(2) = 2^3 + \frac{5}{2}(2^2) = 8 + 10 = 18 \, \text{m} \]