Question

A situation is shown in which two objects A and B start their motion from same point in same direction. The graph of their velocities against time is drawn. u and up are the initial velocities of A and B respectively. T is the time at which their velocities become equal after start of motion. If the value of T is 4s, then the time after which A will meet B is VA Velocity of A B Velocity of B UA T t

Answer 1

8s

Answer 2

To determine the time after which object A will meet object B, we need to analyze their motion based on the given velocity-time graph. Both objects start from the same point in the same direction.

1. Define the equations of motion for A and B:
   Let $u_A$ and $u_B$ be the initial velocities of A and B, respectively, at $t=0$.
   Let $a_A$ and $a_B$ be their constant accelerations (slopes of their respective V-t graphs).

   The velocity equations are:
   $v_A(t) = u_A + a_A t$
   $v_B(t) = u_B + a_B t$

   Since they start from the same point (let's assume $x=0$ at $t=0$), their position equations are:
   $x_A(t) = u_A t + \frac{1}{2} a_A t^2$
   $x_B(t) = u_B t + \frac{1}{2} a_B t^2$

2. Use the condition that their velocities become equal at time T:
   At $t = T$, $v_A(T) = v_B(T)$.
   $u_A + a_A T = u_B + a_B T$
   Rearranging this equation, we get:
   $u_A - u_B = (a_B - a_A) T$
   Given $T = 4s$, so:
   $u_A - u_B = 4(a_B - a_A)$ (Equation 1)

3. Use the condition that A meets B (their positions become equal) at time $t_{meet}$:
   When they meet, $x_A(t_{meet}) = x_B(t_{meet})$.
   $u_A t_{meet} + \frac{1}{2} a_A t_{meet}^2 = u_B t_{meet} + \frac{1}{2} a_B t_{meet}^2$
   Rearrange the terms:
   $(u_A - u_B) t_{meet} + \frac{1}{2} (a_A - a_B) t_{meet}^2 = 0$
   Factor out $t_{meet}$:
   $t_{meet} \left[ (u_A - u_B) - \frac{1}{2} (a_B - a_A) t_{meet} \right] = 0$

   This equation gives two possible solutions for $t_{meet}$:

   • $t_{meet} = 0$: This corresponds to the initial moment when they start from the same point.
   • $(u_A - u_B) - \frac{1}{2} (a_B - a_A) t_{meet} = 0$: This gives the time when they meet again.
     $(u_A - u_B) = \frac{1}{2} (a_B - a_A) t_{meet}$ (Equation 2)

4. Combine Equation 1 and Equation 2:
   Substitute the expression for $(u_A - u_B)$ from Equation 1 into Equation 2:
   $4(a_B - a_A) = \frac{1}{2} (a_B - a_A) t_{meet}$

   From the graph, it is evident that the slopes of the velocity-time graphs are different (A's velocity decreases, B's velocity increases), meaning $a_A \neq a_B$. Therefore, $(a_B - a_A) \neq 0$. We can divide both sides by $(a_B - a_A)$:
   $4 = \frac{1}{2} t_{meet}$
   $t_{meet} = 4 \times 2$
   $t_{meet} = 8s$

Thus, A will meet B after 8 seconds.