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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

2 votes

The man ran 4252.5 meters.

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*(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(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+(1)/(2)at^(2)\\\\x_(2)=4050m+4.5m(m)/(s)*90seconds-(1)/(2)*(0.05(m)/(s^(2)))*(90s)^(2)\\\\x_(2)=4050m+405m-(0.5*0.05(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!

User Peter Stuart
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