Question 453199: A lion begins chasing a gazelle from a distance of 300 m. If the speed of the gazelle is 12kmh and the speed of the lion is 15kmh, how many meters can the gazelle run before being caught?
Thank you!
Answer by marioez(1)   (Show Source):
You can put this solution on YOUR website! Formula
We will use the physics formula d = v.t, which tells you the distance d that you would travel if you were traveling at a constant velocity v, during a given period of time t. So,
d = distance traveled (in meters,m)
v = velocity or speed, which is constant, i.e. does not change (in m/s or mps)
t = time traveled (in seconds, s)
Unit conversion
To match units, we need to convert kmh to m/s or mps,
v(lion) = 15 kmh = 4.17 m/s
v(gazelle) = 12 kmh = 3.33 m/s
Solution
The trick is to realize both animals travel the same time interval t, and equate this time in terms of the distance and velocity of each.
Lets say the distance the lion needs to travel to catch the gazelle is d.
The gazelle will travel 300m less before being caught, or d-300, as shown on the following diagram
<-------------------distance d that lion travels during t------------------>
<-----300m-----><-------distance(d-300) gazelle travels during t------->
Lion....................Gaz........................................................................................Gaz caught
Since both animals travel the same time t, and t = d/v, we can equate both
t(lion) = t(gazelle)
d(lion)/v(lion) = d(gazelle)/v(gazelle)
d/4.17 = (d-300)/3.33
3.33(d/4.17) = d-300
0.8d = d-300
300 = d-0.8d
300 = 0.2d
300/0.2 = d
d = 1500 m distance the lion travels
The gazelle travels d-300, or
d(gazelle) = d(lion)-300 = 1500-300 = 1200 m. answer
Check
t(lion) = t(gazelle)
t(lion) = d(lion)/v(lion) = 1500/4.17 = 360 s
t(gazelle) = d(gazelle)/v(gazelle) = 1200/3.33 = 360 s
It checks, so the answer is 1200 m, the distance the gazelle travels before being caught.
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