Question

1. initially car a is (10.5\text{ m}) ahead of car b. both start moving at time (t = 0) in the same direction along a straight line. the velocity time graph of two cars is shown in figure. the time when the car b will catch the car a, will be :

(a) (t = 21\text{ sec})
(b) (t = 2\sqrt{5}\text{ sec})
(c) (t = 20\text{ sec})
(d) none of these

Explanation:

🆕 New Concept Discovered: Relative Motion and Area under v-t Graph
Using relative displacement and velocity-time graph geometry to find catch-up time.

Step 1: Understand the Motion of Car A

Car A has a constant velocity because its line on the velocity-time graph is horizontal.

$$ v_A(t) = 10\text{ m/s} $$

The displacement of Car A at any time \( t \) is the area under its \( v\text{-}t \) graph:

$$ s_A(t) = 10t $$

Step 2: Determine the Motion of Car B

The velocity of Car B starts from \( 0 \) at \( t = 0 \) and increases linearly. The line makes an angle of \( 45^\circ \) with the vertical velocity axis (\( v \)-axis).

This means the angle with the horizontal time axis (\( t \)-axis) is:

$$ \theta = 90^\circ - 45^\circ = 45^\circ $$

The acceleration \( a_B \) of Car B is the slope of its \( v\text{-}t \) line:

$$ a_B = \tan(45^\circ) = 1\text{ m/s}^2 $$

Since Car B starts from rest (\( u_B = 0 \)), its displacement at any time \( t \) is:

$$ s_B(t) = \frac{1}{2} a_B t^2 = \frac{1}{2} (1) t^2 = 0.5t^2 $$

Step 3: Set up the Catch-up Condition

Initially, Car A is \( 10.5\text{ m} \) ahead of Car B. For Car B to catch Car A, the displacement of Car B must equal the displacement of Car A plus this initial gap:

$$ s_B(t) = s_A(t) + 10.5 $$

Substitute the displacement expressions into the equation:

$$ 0.5t^2 = 10t + 10.5 $$

Step 4: Solve the Quadratic Equation

Multiply the entire equation by \( 2 \) to clear the decimal:

$$ t^2 = 20t + 21 $$

$$ t^2 - 20t - 21 = 0 $$

Factor the quadratic equation:

$$ (t - 21)(t + 1) = 0 $$

This gives two possible solutions for \( t \):

$$ t = 21\text{ s} \quad \text{or} \quad t = -1\text{ s} $$

Since time \( t \) must be positive, we choose:

$$ t = 21\text{ s} $$

Answer:

(A) t = 21 sec