Answer to Question #125721 in Mechanics | Relativity for Lesley

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 30m apart

B)Particles meet

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

Answer 1

In order to solve this problem, we have to remember about two things:

1. If the point moves with some acceleration, then the vector of the point coordenates depends from time like (1)
2. In the same conditions the vector of a velocity depends on time like (2)

$\vec {OP} = \vec V_pt + \frac{\vec a_pt^2}{2}$(1) where $\vec V_p, \vec a_p$ the speed and acceleration of the point P, O - the start point

$\vec V_p = \vec V_o + \vec a_p t$ (2) where $\vec V_o$ the vector of initial velocity.

I had found the vector of distance between two points moving toward each other($\vec {PQ}$ ).

Problem A:

Because the vectors $\vec{PQ},\vec{MP},\vec{QN}$ is collinearly, we can think about objects like about scalars. In this problem we must say: $PQ$ = 30. I have considered vector: $\vec{PQ} = \vec{MN}-\vec{MP}-\vec{NQ} = \vec{MN} - \vec V_Pt - \frac{\vec a_Pt^2}{2} - \vec V_Qt - \frac{\vec a_Qt^2}{2}$

The equation in scalars will be next:

$PQ = MN - V_Pt-\frac{a_Pt^2}{2}-V_Qt-\frac{a_Qt^2}{2}.$

After substitution the condition values of the problem:

$30 = 51-5t-\frac{1}{2}t^2-6t-\frac{3}{2}t^2$ or $t^2+\frac{11}2{}t-\frac{21}{2}=0$

This qudratic equation is easy to solve:

$t_1 = -7, t_2 = \frac{3}{2}$ the last root is right.

Problem B:

The solution of the problem is the solution of problem A, we just say that $PQ=0$

Problem C:

$V_P+a_Pt = \frac{3}{4}(V_Q+a_Qt)$

This is the linear equation. The solution of the one is:

$t = \frac{V_P-\frac{3}{4}V_Q}{\frac34a_Q-a_P} = \frac{2}{5}s$