# Thread: Velocity of a bullet

1. ## Velocity of a bullet

Here is another one I am working on. I need help finding the linear equation, if I could, please. Once I have the equation, I can figure the rest out, I believe.

Velocity of a bullet.
A gun is fired straight upward. The bullet leaves the gun at 100 feet per second (time t=0). After 2 seconds the velocity of the bullet is 36 feet per second. There is a linear equation that gives the velocity v in terms of the time t. Find the equation and find the velocity after 3 seconds.

Alright. First, read the question and collect all information :

- it is a linear equation, so it has the form $y = mx + p$ (I believe it is useless to check if the line is parallel to the y-axis, that would be boring )
- we are talking about velocity. Therefore, we are talking of feet per second.
- the bullet leaves the gun at 100 feet per [1] second. Doesn't this remind us of something ? Yes, I guess that would be the slope of the line : if the bullet goes up 100 feet (y-axis) for 1 second (x-axis), then the slope will be $rise/run = 100 / 1 = 100$. Therefore, in the equation, $m = 100$.
- After 2 seconds the velocity of the bullet is 36 feet per second. Hmm ... maybe this could actually be a point of the linear equation. Therefore, the point $(2 ; 36)$ belongs to the equation.

Good. We know that $y = 100x + p$, and that $(2 ; 36)$ is a point of the equation. But we are still missing $p$ to complete the equation. Let's try to substitute the values of the point that we have into the current equation to find $p$ (since the point belongs to the equation) :

$36 = 100 * 2 + p$
$36 = 200 + p$
$36 - 200 = p$
$p = -164$

We found $p$ ! Now our equation is complete : $y = 100x - 164$

Let's conclude : the velocity $y$ of the bullet (in feet) at time $x$ (in seconds) is given by the linear equation $y = 100x - 164$
Note that I used $x$ and $y$ to denote the velocity and time. You can change following the question, therefore the equation is :

$v = 100t - 164$

The second question should be easy now

3. Thank you for your quick reply. I am calling it a night, but will look over your notes in the morning and apply the information you have supplied.

I sincerely appreciate the help.

4. Ok, I have worked though your figures. Thank you again.

So, if v=100t-164 for the main equation, this is what I got for the velocity of the bullet after 3 seconds:

v=100t-264

I reached this by doing 36=100(3)+b
(I am familiar with the linear equation of y=mx+b, so I followed that).

So, 36=100(3)+b, to 36=300+b, to -264=b, or b=-264

I substituted the letter v for y, and the letter t for x.

Would you concur?

Nah, the equation is $v = 100t - 164$, and that's all. Now you want the velocity "v" for a time t = 3. Therefore, you substitute 3 into "t" in the equation :

$v = 100 * 2 - 164$
$v = 200 - 164$
$v = 36$

Therefore, at 3 seconds, the velocity of the bullet is 36 feet per second.

6. Originally Posted by Bacterius
Nah, the equation is $v = 100t - 164$, and that's all. Now you want the velocity "v" for a time t = 3. Therefore, you substitute 3 into "t" in the equation :

$v = 100 * 2 - 164$
$v = 200 - 164$
$v = 36$

Therefore, at 3 seconds, the velocity of the bullet is 36 feet per second.
Wouldn't substituting 3 for 't' mean v=100*3-164?

So v=300-164, to v=136?

Therefore, at 3 seconds, the velocity of the bullet is 136 feet per second.

As for clarifying the other part of the equation, thank you very much. I sincerely appreciate the help and clarification.