The problem describes two particles moving along the same straight line. From the given $v-t$ graph, we can determine the acceleration of each particle based on the slope of their respective lines. Let the intersection point where both particles have the same velocity occur at time $t = 4$ s with a velocity $v_0$.
For Particle 1: It starts from rest ($u_1=0$) at $t=0$ and reaches velocity $v_0$ at $t=4$.
$$ a_1 = \frac{\Delta v}{\Delta t} = \frac{v_0 – 0}{4 – 0} = \frac{v_0}{4} $$For Particle 2: It starts from rest ($u_2=0$) at $t=3$ and reaches velocity $v_0$ at $t=4$.
$$ a_2 = \frac{\Delta v}{\Delta t} = \frac{v_0 – 0}{4 – 3} = v_0 $$Comparing the two accelerations:
$$ \frac{a_1}{a_2} = \frac{v_0 / 4}{v_0} = \frac{1}{4} \implies a_2 = 4a_1 $$The chase ends when Particle 2 catches up to Particle 1. Since they are moving on the same straight line and started from the same reference point, their displacements must be equal at the moment they meet.
Let $t$ be the total time elapsed from $t=0$ when the meeting occurs.
- Particle 1 travels for time $t$.
Displacement $S_1 = u_1 t + \frac{1}{2}a_1 t^2 = \frac{1}{2}a_1 t^2$ - Particle 2 starts at $t=3$, so it travels for time $(t-3)$.
Displacement $S_2 = u_2(t-3) + \frac{1}{2}a_2 (t-3)^2 = \frac{1}{2}a_2 (t-3)^2$
Equating the displacements ($S_1 = S_2$):
$$ \frac{1}{2}a_1 t^2 = \frac{1}{2}a_2 (t-3)^2 $$Substitute $a_2 = 4a_1$ into the equation:
$$ \frac{1}{2}a_1 t^2 = \frac{1}{2}(4a_1)(t-3)^2 $$ $$ t^2 = 4(t-3)^2 $$Taking the square root of both sides:
$$ t = \pm 2(t-3) $$Case 1: $t = 2(t-3)$
$$ t = 2t – 6 \implies t = 6 \text{ s} $$Case 2: $t = -2(t-3)$
$$ t = -2t + 6 \implies 3t = 6 \implies t = 2 \text{ s} $$The solution $t=2$ s is not valid because Particle 2 does not even start moving until $t=3$ s. Thus, the only physical solution is $t=6$ s.