Question

Two particles are at the same point at the same time, moving in the same direction. Particle A has an initial velocity of 7.6 m/s and an acceleration of 2.6 m/s 2 . Particle B has an initial velocity of 3.4 m/s and an acceleration of 5.8 m/s 2 . At what time will B pass A?

Answer 1

Answer:

2.625 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

For particle A

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=7.6\times t+\frac{1}{2}\times 2.6\times t^2\\\Rightarrow s=7.6t+1.3t^2$

For particle B

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=3.4\times t+\frac{1}{2}\times 5.8\times t^2\\\Rightarrow s=3.4t+2.9t^2$

Now, the displacement should be same if they are about to cross each other.

$7.6t+1.3t^2=3.4t+2.9t^2\\\Rightarrow 7.6t-3.4t=2.9t^2-1.3t^2\\\Rightarrow 4.2t=1.6t^2\\\Rightarrow t=\frac{4.2}{1.6}\\\Rightarrow t=2.625\ s$

At 2.625 seconds the particles would have covered equal distances after which B would pass A.