Question #125721

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

Expert's answer

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

- If the point moves with some acceleration, then the vector of the point coordenates depends from time like (1)
- 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"

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