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1st Equation of Motion

Two particles...

Question

Two particles $P$ and $Q$ move in a straight line $A B$ towards each other. $P$ starts from $A$ with velocity $u_{1}$ , and an acceleration $a_{1}$ , $Q$ starts from $B$ with velocity $u_{2}$ and acceleration $a_{2}$ .They pass each other at the midpoint of $A B$ and arrive at the other ends of $A B$ with equal velocities

They meet at midpoint at time

t = \frac{2 (u_{2} - u_{1})}{a_{1} - a_{2}}

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The length of path specified i.e.

A B

l = \frac{4 (u_{2} - u_{1}) (a_{1} u_{2} - a_{2} u_{1})}{(a_{1} - a_{2})^{2}}

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They reach the other ends of

A B

with equal velocities if

(u_{2} + u_{1}) (a_{1} - a_{2}) = 8 (a_{1} u_{2} - a_{2} u_{1})

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They reach the other ends of

A B

with equal velocities if

(u_{2} - u_{1}) (a_{1} + a_{2}) = 8 (a_{2} u_{1} - a_{1} u_{2})

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Solution

The correct options are
A They meet at midpoint at time $t = \frac{2 (u_{2} - u_{1})}{a_{1} - a_{2}}$
B The length of path specified i.e. $A B$ is $l = \frac{4 (u_{2} - u_{1}) (a_{1} u_{2} - a_{2} u_{1})}{(a_{1} - a_{2})^{2}}$
D They reach the other ends of $A B$ with equal velocities if

$(u_{2} + u_{1}) (a_{1} - a_{2}) = 8 (a_{1} u_{2} - a_{2} u_{1})$

Since both particles traveled same distance,
$u_{1} t + \frac{1}{2} a_{1} t^{2} = u_{2} t + \frac{1}{2} a_{2} t^{2} \Rightarrow t = 0 a n d t = \frac{2 (u_{2} - u_{1})}{a_{1} - a_{2}}$
$u_{r e l} = u_{1} + u_{2} a_{r e l} = a_{1} + a_{2} s = \frac{l}{2} = (u_{1} + u_{2}) \times \frac{2 (u_{2} - u_{1})}{a_{1} - a_{2}} + \frac{1}{2} (a_{1} + a_{2}) \times \frac{4 {(u_{2} - u_{1})}^{2}}{{(a_{1} - a_{2})}^{2}} \Rightarrow l = \frac{4 (u_{2} - u_{1}) (a_{1} u_{2} - a_{2} u_{1})}{{(a_{1} - a_{2})}^{2}}$

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