Janiah Parks

2022-09-04

A car leaves an airport going 84 km/hr. 5 hours later, an airplane going 705 km/hr departs. How do you find the distance when the airplane will catch up with the car?

aveanoon

Beginner2022-09-05Added 8 answers

Call the distance you need $d={x}_{f}-{x}_{i}$ and the time the airplane needs T.

In general velocity will be $v=\frac{{x}_{f}-{x}_{i}}{t}$

if ${x}_{i}=0$ (the airport, for both) we can write:

${x}_{f}=vt$ (1)

so for the car and airplane $x}_{f$ has to be the same; so:

$x}_{fcar}={x}_{fair$

The cat will travel 5+T hours and the airplane T hours:

$84(5+T)=705T$

$425+84T=705T$

$T=\frac{425}{621}=0.684h$

using this into the expression for, say, the car we get:

${x}_{f}=84(5+T)=477.5km$

The only thing is that you are given velocities without considering that the car and the airplane has to start from zero and accelerate...the passengers probably would not survive going from zero to 705km/hr instantly!

In general velocity will be $v=\frac{{x}_{f}-{x}_{i}}{t}$

if ${x}_{i}=0$ (the airport, for both) we can write:

${x}_{f}=vt$ (1)

so for the car and airplane $x}_{f$ has to be the same; so:

$x}_{fcar}={x}_{fair$

The cat will travel 5+T hours and the airplane T hours:

$84(5+T)=705T$

$425+84T=705T$

$T=\frac{425}{621}=0.684h$

using this into the expression for, say, the car we get:

${x}_{f}=84(5+T)=477.5km$

The only thing is that you are given velocities without considering that the car and the airplane has to start from zero and accelerate...the passengers probably would not survive going from zero to 705km/hr instantly!

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