Need some help with this one please
A hicker walking at a constant rate of 4 miles per hour is passed by a cyclist traveling in the same direction along the same path at a costant rate of 20 miles per hour. The cyclist stops to wait for the hicker 5 minutes after passing her, while the hicker continues at her constant rate. How many minutes must the cyclist wait until the hicker catches up?
The answers is 20
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Rate problem
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lukeposada
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cyclist rate is 20 mile per 1 hour = 1/3 mile per 1 minute
therefore, in 5 minutes he'll pass 5/3 mile
walker rate is 4 mile per 1 hour = 1/15 mile per 1 minute
therefore:
it will take the walker t=(5/3)/(1/15)=25 minutes to get to the same point as the cyclist
so, the cyclist will have to wait for him for 20 minutes
hope it helps
therefore, in 5 minutes he'll pass 5/3 mile
walker rate is 4 mile per 1 hour = 1/15 mile per 1 minute
therefore:
it will take the walker t=(5/3)/(1/15)=25 minutes to get to the same point as the cyclist
so, the cyclist will have to wait for him for 20 minutes
hope it helps
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barcebal
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Here's how I try to look at rate problems where two objects travel in the SAME direction at different rates.
The hiker is going 4 MPH and the cycler is going 20 MPH. So when they cross, the cycler begins to exceed the distance traveled of the hiker by 16 miles every hour (20-4=16).
Or in other words the difference in their individual rates is the rate of the gain of the faster object gains on the slower object. It makes sense because after one hour, what has the cycler gained on the hiker? Well the hiker moved 4 miles and the cyclist moved 20, so 16. After 2 hours? Hiker is at 8 miles and the cyclist is at 40, so 32.
So, once again, the rate of the GAIN is the difference in the individual rates. This is the key to understanding the question.
So you can use the rate we just found (16 mph is the rate of the gain of the cycler on the hiker) in the formula D=rt to figure out how much the cycler has EXCEEDED the hiker after 5 minutes.
D=(16)*(5/60)=the distance the cycler has EXCEEDED hiker after 5 minutes.
We solve to find that the distance is 4/3 of a mile.
Now we just have to figure out how long it will take the hiker to move 4/3 of a mile while the cycler doesn't move.
4/3 mile =4 mph * (t)
t=1/3
1/3 represents 1/3 of an hour which is 20 minutes.
I hope this helps.
The hiker is going 4 MPH and the cycler is going 20 MPH. So when they cross, the cycler begins to exceed the distance traveled of the hiker by 16 miles every hour (20-4=16).
Or in other words the difference in their individual rates is the rate of the gain of the faster object gains on the slower object. It makes sense because after one hour, what has the cycler gained on the hiker? Well the hiker moved 4 miles and the cyclist moved 20, so 16. After 2 hours? Hiker is at 8 miles and the cyclist is at 40, so 32.
So, once again, the rate of the GAIN is the difference in the individual rates. This is the key to understanding the question.
So you can use the rate we just found (16 mph is the rate of the gain of the cycler on the hiker) in the formula D=rt to figure out how much the cycler has EXCEEDED the hiker after 5 minutes.
D=(16)*(5/60)=the distance the cycler has EXCEEDED hiker after 5 minutes.
We solve to find that the distance is 4/3 of a mile.
Now we just have to figure out how long it will take the hiker to move 4/3 of a mile while the cycler doesn't move.
4/3 mile =4 mph * (t)
t=1/3
1/3 represents 1/3 of an hour which is 20 minutes.
I hope this helps.
