Answer to Question #206106 in Mechanics | Relativity for marie

Question

Balls A and B accelerate from rest towards each other at 2 m/s² and 3 m/s² respectively. Given that the initial distance between the balls is 1500 m, determine:

i. the time taken for the balls to meet.

ii. the distance covered by ball A when the balls meet.

iii. the distance travelled by ball B when the balls meet.

Answer 1

Gives

$a_1=2m/sec$

$a_2=3m/sec^2$

Distance (s)=$1500m$

Ball A

Time take t₁

$s=ut+\frac{1}{2}at^2$

s=1500-x

$u=0$

$t_1=\sqrt\frac{2s}{a_1}$

$t_1=\sqrt\frac{2(1500-x)}{2}$

Second ball

s=x

$t_2=\sqrt\frac{2\times x}{a_2}$

$t_1=t_2$

$\sqrt\frac{2(1500-x)}{2}=\sqrt\frac{2(x)}{3}$

Both side squre take

$1500-x=\frac{2x}{3}$

$4500-3x=2x$

$5x=4500$

$x=900m$

Both ball are meet same time

$t_1=t_2=t=\sqrt\frac{2\times900}{3}=\sqrt{600}sec$

Part (b)

Distance covered by ball A

$x'=1500-900=600m$

Part( c)

Dixtance covered ball B

$x=x''=900m$