Answer to Question #125969 in Physics for Lesley Smarts

Question

Two particles P and Q move towards each other along a straight line MN, 51 meters long. P starts from M with velocity 5ms^-1 and constant acceleration of 1ms^-2. Q starts from N at the same time with velocity 6ms^-1 and at a constant acceleration of 3ms^-2.

Find the time when the :

A) particles are 30 metres apart

B)Particles meet

C) velocity of P is ¾ of the velocity of Q

Answer 1

(a)

$s_1=v_{01}t+\frac{a_1t^2}{2}$ and $s_2=v_{02}t+\frac{a_2t^2}{2}$

$51-(s_2+s_1)=30 \to s_2+s_1=21$

$v_{01}t+\frac{a_1t^2}{2}+v_{02}t+\frac{a_2t^2}{2}=21 \to 5t+\frac{t^2}{2}+6t+\frac{3t^2}{2}=21 \to$

$t=1.5s$

(b)

$5t+\frac{t^2}{2}+6t+\frac{3t^2}{2}=51 \to t=3s$

(c)

$v_1=v_{01}+a_1t=5+t$ and $v_2=v_{02}+a_2t=6+3t$

$v_1=\frac{3}{4}v_2\to 5+t=4.5+2.25t\to t=0.4s$