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AfilCa [17]
4 years ago
13

Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia

lly 5.0 km east of the flagpole and is running with a constant velocity of 8.0 km/h due west. What will be the distance of the two runners from the flagpole when their paths cross
Physics
1 answer:
Eduardwww [97]4 years ago
7 0

Answer:

From the data we know that runner A and runner B are 11 km apart from the start because (6+5) km

So the runner from the east direction has distance as unknown km, rate= 9 k/h ; time= d/r=x/9 hr

So runner towards the west will be

distance = 11-x, rate= 8 k/h, time = d/r = (11-x)/8

So equating east and west time we have

x/9= (11-x)/8

8x=99-9x

17x=99

x=5.92 km

That is the distance covered by runner towards the east and he will meet the runner toward the west at

6-5.92=0.08 km west of the flagpole.

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The original kinetic energy will be 0 J and the final kinetic energy will be 7500 J and the amount of work utilized will be similar to the final kinetic energy i.e., 7500 J.

<u>Explanation:</u>

As it is known that the kinetic energy is defined as the energy exhibited by the moving objects. So the kinetic energy is equal to the product of mass and square of the velocity attained by the car. Thus,

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So the initial kinetic energy will be the energy exerted by the car at the initial state when the initial velocity is zero. Thus the initial kinetic energy will be zero.  

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\text {Kinetic energy}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 600 \times 5 \times 5 = 7500 J

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