Question

Two particles are moving along positive x-axis, Particle-1 is ahead of particle-2. The initial velocities of particle-1 and particle-2 are 5m/s and 10m/s respectively. The acceleration of particle-1 and particle-2 are 5m/s2 and 8m/s2 and are along positive x-axis. The initial gap between the particles is 16 metres. The time as which they meet one another is​

Answer 1

Answer:

1sec

Explanation:

For particle-1 displacement(S)=5t-5t^2/2

For particle-2 displacement(S`)=10t+4t^2

S+S`= 16m given in question

upon substituting we get t=0.97s or t=-10s

so answer is t=1sec approximately

Answer 2

Answer: The correct answer is after 2s, 1 and 2 will meet one another.

Explanation:

• Let both the particles meet one another at time t.
• The distance covered by particle 1 is given by the equation:
• $s = u_1t+\frac{1}{2}a_1t^2\\s = 5t+\frac{1}{2}5t^2---- > 1$
• Since particle 2 meets 1 at a common point and the gap between them was 16m, So the distance covered by the particle 2:
• $s +16 = u_2t+\frac{1}{2}a_2t^2\\s = 10t+\frac{1}{2}8t^2 - 16\\s = 10t+4t^2 - 16--- > 2$
• From Equations 1 and 2
• $5t+\frac{5}{2}t^2=10t+4t^2-16\\10t+5t^2=20t+4t^2-32\\3t^2+10t-32=0\\therefore \hspace{2} t = \frac{-10\pm\sqrt{100-(-4\times3\times32)}}{2\times3}\\t = \frac{-10\pm22}{6}\\t = 2s \hspace{5} t = -5.33s$
• Therefore t = 2s