Question

Initially car A is 10.5 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 sec
• B. t=$2\sqrt{5}$ sec
• C. 20 sec
• D. None of these

Answer 1

Answer: (A)

t=21 sec

Answer 2

Understanding the Problem:

We need to find the time when car B catches up to car A. This happens when their positions are equal.

Approach:

1. Determine the equations of motion for both cars.
2. Set the position equations equal to each other.
3. Solve for time (t).

Detailed Solution:

• Car A: Moves with constant velocity $v_A = 10 \, \text{m/s}$. Its initial position is $x_A(0) = 10.5 \, \text{m}$. Therefore, its position as a function of time is:

  $x_A(t) = x_A(0) + v_A t = 10.5 + 10t$

• Car B: Starts from rest and accelerates. From the graph, we can see that $v_B(t) = t$. Therefore, its acceleration $a_B = 1 \, \text{m/s}^2$. Initial position $x_B(0) = 0$. Its position as a function of time is:

  $x_B(t) = x_B(0) + v_B(0)t + \frac{1}{2}a_B t^2 = \frac{1}{2}t^2$

• Equating positions to find the time when car B catches car A:

  $x_A(t) = x_B(t)$ $10.5 + 10t = \frac{1}{2}t^2$ Multiplying by 2: $21 + 20t = t^2$ Rearranging: $t^2 - 20t - 21 = 0$

• Solving the quadratic equation:

  Using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $t = \frac{20 \pm \sqrt{(-20)^2 - 4(1)(-21)}}{2}$ $t = \frac{20 \pm \sqrt{400 + 84}}{2}$ $t = \frac{20 \pm \sqrt{484}}{2}$ $t = \frac{20 \pm 22}{2}$

  This gives two possible solutions: $t = 21 \, \text{s}$ or $t = -1 \, \text{s}$. Since time cannot be negative, the correct solution is $t = 21 \, \text{s}$.