A man jogs at a speed of 1.1 m/s. His dog waits 1.6 s and then takes off running at a speed of 4.1 m/s to catch?
Please help me understand my physics!
Update:A man jogs at a speed of 1.1 m/s. His dog
waits 1.6 s and then takes off running at a
speed of 4.1 m/s to catch the man.
How far will they have each traveled when
the dog catches up with the man?
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Verified answer
Distance = velocity * time
Man
d = 1.1t
Dog
d = 4.1(t -1.6) = 4.1 t - 6.56
Since they both run the same distance set the equations equal
1.1t = 4.1t - 6.56
Subtract 1.1t from both sides
0 = 3t - 6.56
Add 6.56 to both side
6.56 = 3t
3t = 6.56
Divide both sides by 3
t = 2.1866666667
Distance travel
d = 1.1* 2.1866666667
d = 2.4053333334
They traveled 2.4 meters
m09262013
For a visual aid try graphing this out. Use the x and y axis for meters and seconds.
Plotting the movement and elasped time for each you will be looking for the 'intersection' of those. That will give you how far the man and dog ran before meeting. Since the dog is run apprx. 4 times as fast as the man his line should go on much farther than the man for the same amount of time.
The man will have jogged 2.4053 m, after .58665 seconds (2.18665 overall).