Acceleration with Velocity Problem.

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SUMMARY

The problem involves a boy running at a constant velocity of 3.0 m/s [E] and his father, who starts from rest and accelerates at 1.5 m/s². To determine how long it takes for the father to catch up to the son, one must set the distances equal: d_father = d_son. The correct time for the father to catch his son is 4 seconds, not 2 seconds, as the initial calculation only considered the time to reach the same speed without accounting for the distance already covered by the son.

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A boy is running a a constant velocity of 3.0 m/s [E] and passes his father who is standing still. The father immediately starts to chase his son with a constant acceleration of 1.5m/s at the instant his that his son reachers him

How long does it take the father to catch his son?

So to find time. I did

T= vf - vi/aav

Vf is final velocity (3.0-0)
Vi is initial velocity (0)
Aav is average velocity (1.5)

I got 2 Seconds? But the correct answer is 4? How?
 
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You calculated the time it would take the father to reach the same speed as the son. But he will still be some way behind the son.
 
haruspex said:
You calculated the time it would take the father to reach the same speed as the son. But he will still be some way behind the son.

so what do I do though...
 
Write equations for how far each will travel in time t after the son passes the father.
 
at the instant his father catches his son, they all have the same distance from since the son passed his father, hence:
d_{father}=d_{son}

continue here
 

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