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

Step 1: Understanding the Concept:
When acceleration is a function of time, we cannot use the standard equations of motion. Instead, we must use calculus to integrate acceleration to find velocity, and then integrate velocity to find displacement.
: Key Formula or Approach:
1. Velocity \( v = \int a \, dt \).
2. Displacement \( s = \int v \, dt \).
Step 2: Detailed Explanation:
Given acceleration \( a = \frac{dv}{dt} = 6t + 5 \).
Integrating with respect to time to find velocity \( v \):
\[ v = \int (6t + 5) \, dt = \frac{6t^2}{2} + 5t + C_1 = 3t^2 + 5t + C_1 \]
Since the point starts from rest, at \( t = 0 \), \( v = 0 \). This gives \( C_1 = 0 \).
\[ v = 3t^2 + 5t \]
Now, integrate the velocity to find the displacement \( s \):
\[ s = \int (3t^2 + 5t) \, dt = \frac{3t^3}{3} + \frac{5t^2}{2} + C_2 = t^3 + 2.5t^2 + C_2 \]
Since it starts from the origin, at \( t = 0 \), \( s = 0 \). This gives \( C_2 = 0 \).
The displacement equation is:
\[ s = t^3 + 2.5t^2 \]
Substitute \( t = 2 \text{ s} \) into the equation:
\[ s = (2)^3 + 2.5(2)^2 \]
\[ s = 8 + 2.5(4) \]
\[ s = 8 + 10 = 18 \text{ m} \]
Step 3: Final Answer:
The distance covered in 2 s is 18 m.