Your friend is traveling at a constant 30.0m/s and has a 1600m head start. How many minutes will it take you to catch them if you are traveling at a constant 50.0m/s?

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Your friend is traveling at a constant 30.0m/s and has a 1600m head start. How many minutes will it take you to catch them if you are traveling at a constant 50.0m/s?

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Answer 1 · 2015-11-09T03:20:48

80 seconds

Explanation:

By defining $t$ $t$ as the time it will take to you and your friend to be in the same position $x$ $x$ ; $x_{0}$ $x_0$ is the starting position and using the equation of motion $x = x_{0} + v t$ $x=x_0 + vt$ you have:
$x = 1600 + 30 \cdot t$ $x = 1600 + 30*t$
$x = 0 + 50 \cdot t$ $x = 0 + 50*t$
Since you want the moment when both are in the same position, that is the same $x$ $x$ , you make both equations equal.
$1600 + 30 \cdot t = 50 \cdot t$ $1600 + 30*t = 50*t$
And solving for $t$ $t$ to know the time:
$t = \frac{1600 m}{50 \frac{m}{s} - 30 \frac{m}{s}} = \frac{1600 m}{20 \frac{m}{s}} = 80 s$ $t = (1600 m)/(50 m/s -30 m/s) = (1600 m)/(20 m/s) = 80 s$

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Your friend is traveling at a constant 30.0m/s and has a 1600m head start. How many minutes will it take you to catch them if you are traveling at a constant 50.0m/s?

1 Answer

Explanation:

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