# Thread: A simple problem, two factors

1. ## A simple problem, two factors

Rehearsing a bit of math I come across this problem I've got a hard time to get a grip on.
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A person shoots at a barrel, two seconds later he hears the bullet hit the barrel. The speed of sound is 340m/s and the average speed of the bullet is 650m/s.

What is the distance between the shooter and the barrel?
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Edit: Maybe the distance can be expressed as x and ...?

2. Hello, λιεҗąиđ€ŗ!

A person shoots at a barrel.
Two seconds later he hears the bullet hit the barrel.
The speed of sound is 340m/s and the average speed of the bullet is 650m/s.

What is the distance between the shooter and the barrel?
We know that: .$\displaystyle \text{Distance } = \text{Speed }\times \text{Time}\quad\Rightarrow\quad T \:=\:\frac{D}{S}$

Let $\displaystyle x$ = distance from the shooter to the barrel (in meters).

. . At 650 m/s, the bullet takes: .$\displaystyle \frac{x}{650}$ seconds to reach the barrel.

. . At 340 m/s, the sound takes: .$\displaystyle \frac{x}{340}$ seconds to reach the shooter.

Since the total time is 2 seconds, we have: .$\displaystyle \frac{x}{650} + \frac{x}{340} \:=\:2$

. . Now solve for $\displaystyle x$ . . .

3. Could the two x's be merged? Only way I see to make some progress?

Btw, why does my text come out as red? lol.

Edit: On second thought, should this thread be closed and a new one created for purely solving the equation?

4. Heloo, λιεҗąиđ€ŗ!

Could the two x's be merged? . . . . Of course!
Do you need an Algebra review?

Multiply through by the LCD: $\displaystyle 2210$

. . $\displaystyle 34x + 65x \;=\;4420\quad\Rightarrow\quad 99x \;=\;4420$

Therefore: .$\displaystyle x \;=\;\frac{4420}{99} \;=\;44.\overline{64}$