Another kinematics easy problem

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SUMMARY

The discussion centers on a kinematics problem involving two ships: one belonging to the evil Dr. X and the other to a hero pursuing her. The average speed of Dr. X's ship, denoted as (v_av)_x, is related to the hero's speed, (v_av)_H, through the equation (v_av)_H = 1.3125(v_av)_x. The calculations are based on the distances traveled by both ships over their respective times of 8.4 hours and 6.4 hours. The approach taken by the participants is confirmed as correct, providing a clear expression for the hero's speed in relation to Dr. X's speed.

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Omid
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Dear PF,
Hera is a problem I am trying to solve.

The evil Dr.X secretly leaves Space port L4 in a warship capable of traveling at an average speed of (v_av)_x.
Two hours later her escape is noticed and our hero blasts off after her at a speed that will effect rendezvous in 6.4 hours. Write an expression for his speed, (v_av)_H, in terms of hers, (v_av)x.

I assumed the place they meet is where L (the distance traveled) is equal for both then :
(1) (v_av)_x = L / (8.4)
(2) (v_av)_H = L / (6.4)
so (v_av)_H = 1.3(v_av)x
If I'm making a mistake please let me know.
I don't know why the writer of this book hasn't included
an answer section :mad:

Thanks
 
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Looks good.
 
for sharing this problem with us, and for showing your work! Your approach seems correct to me. In order to find the expression for (v_av)_H in terms of (v_av)_x, we can use the fact that both the warship and the hero's ship travel the same distance, L, in their respective times of 8.4 hours and 6.4 hours. This means that the average speeds can be expressed as L/8.4 and L/6.4 for the warship and the hero's ship, respectively. Since we want to express (v_av)_H in terms of (v_av)_x, we can set these two expressions equal to each other and solve for (v_av)_H. This gives us (v_av)_H = (8.4/6.4)(v_av)_x = 1.3125(v_av)_x. So your answer of (v_av)_H = 1.3(v_av)_x is correct! Keep up the good work!
 

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