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2 years ago

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A man runs at a velocity of 4.5 m/s for 15.0 min. When going up an increasingly steep hill, he slows down at a constant rate of

0.05 m/s^2 for 90.0s and comes to a stop.
How far did he run?

1 answer:

madreJ [45]2 years ago

4 0

The man ran <u>4252.5 meters.</u>

Why?

To solve the problem, we need to divide the exercise into two movements, the first on while the was running at 4.5 m/s for 15 min, and then, while he was slowing down (going up because of the hill).

First movement: Running at 4.5 m/s for 15 min.

We need convert from minutes to seconds,

$1min=60seconds\\\\15min*\frac{60seconds}{1min}=900seconds$

Now, calculating the distance covered for the first movement, we have:

$x_{1}=0+v_{1}*t_{1}\\\\x_{1}=4.5\frac{m}{s}*900s=4050m$

So, we know that the man covered 4050m for the first movement, it will be our initial position for the second movement.

Second movement: acceleration -0.05m/s^2 (because he's slowing down) for 90 seconds, at 4.5m/s.

$x_{2}=x_{1}+v_{1}*t+\frac{1}{2}at^{2}\\\\x_{2}=4050m+4.5m\frac{m}{s}*90seconds-\frac{1}{2}*(0.05\frac{m}{s^{2}})*(90s)^{2}\\\\x_{2}=4050m+405m-(0.5*0.05\frac{m}{s^{2}}*8100s^{2})=4050m+405m-202.5m\\\\x_{2}=4252.5m$

Hence, we have that he ran 4252.5 m.

Have a nice day!

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