Question

Two particles are moving along x-axis. Particle-1 is 40 m behind Particle-2. Particle-1 starts with velocity12 m//s and acceleration 4 m//s^2 both in positive x-direction. Particle-2 starts with velocity 4 m//s and acceleration 12 m//s^2 also in positive x-direction. Find (a) the time when distance between them is minimum. (b) the minimum distacne between them.

Answer 1

Given:

Two particles are moving along x-axis.

Particle-1 is 40 m behind Particle-2.

Particle-1 starts with velocity 12 m//s and acceleration 4 m//s^2 both in positive x-direction.

Particle-2 starts with velocity 4 m//s and acceleration 12 m//s^2 also in positive x-direction

To Find:

The time when distance between them is minimum.

The minimum distance between them.

Solution:

Displacement of particle 1, S1

• V1 = 12m/s
• a1 = 4m/s²
• S1 = V1t + 0.5 x at² = 12t + 2t²

Displacement of particle 2, S2

• V2 = 4m/s
• a2 = 12m/s²
• S2 = 4t + 6t² + 40

Distance between the two particles:

• S2 - S1 = 4t² - 8t + 40 = S

for minimum distance,

• S' = 0
• 8t - 8 = 0 , t = 1

Therefore time when distance between them is minimum is,

• t = 1sec

• Minimum distance = 4 - 8 + 40 = 36m