Question

Two cars A and B are starts moving from rest. Car A travels at a constant velocity of 20 m/s and car B moving with a constant acceleration of 4 m/s$^2$. After how much time B will overtake A?

• A. 5 s
• B. 10 s
• C. 15 s
• D. 20 s
• E. 25 s

Hint:

Always set up equations for position as a function of time for all moving objects. Overtaking or meeting simply means setting their position equations equal to each other.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a 1D kinematics problem involving two bodies. 'Overtaking' means that both cars have covered the exact same displacement from their starting point at the same time \(t\).
Step 2: Key Formula or Approach:
For Car A (constant velocity): \(S = vt\)
For Car B (constant acceleration from rest): \(S = ut + \frac{1}{2}at^2\)
Equate the distances \(S_A = S_B\) to find time \(t\).
Step 3: Detailed Explanation:
Let the time taken to overtake be '\(t\)' seconds.
For Car A:
Velocity, \(v_A = 20\) m/s
Displacement, \(S_A = v_A \cdot t = 20t\)
For Car B:
Initial velocity, \(u_B = 0\) m/s
Acceleration, \(a_B = 4\) m/s\(^2\)
Displacement, \(S_B = u_B \cdot t + \frac{1}{2} a_B \cdot t^2 = 0 + \frac{1}{2}(4)t^2 = 2t^2\)
Condition for overtaking:
Both cars cover the same distance.
\[ S_A = S_B \]
\[ 20t = 2t^2 \]
Assuming \(t>0\) (time after starting), we can divide by \(2t\):
\[ 10 = t \]
So, \(t = 10\) seconds.
Step 4: Final Answer:
Car B will overtake Car A after 10 seconds.