I do not understand how you can subtract [math]16.1ft/s^2[/math] from [math]40ft/s[/math]...according to a solution to an answer I am looking at this would result in just 23.9 of an unspecified unit.
To extract the part of the problem without bothering with the entire context I am looking at, its saying that
[math](16.1 ft/s^2)t^2=(40 ft/s)t-(40 ft/s)+(16.1 ft/s^2)t^2-(32.2ft/s^2)t+(16.1ft/s^2)[/math][math]7.8t = 23.9[/math][math]t=23.9/7.8[/math][math]t=3.06s[/math]
What is happening when you subtract ft/s^2 from ft/s to result in 7.8t=23.9? Thats all I really have a problem with. If I know that I can try to understand the rest of the problem.
I understand arithmetic can be done with units so I have a superficial understanding of something like:
[math]\frac{(m/s)^2}{m/s^2}=\frac{m^2/s^2}{m/s^2}=\frac{m^2}{s^2} \div \frac{m}{s^2}=\frac{m^2}{s^2} \times \frac{s^2}{m}=m[/math]
but if I were to do the same according to my example I refer to, assuming its not wrong, I'd have to try something like this:
[math]\frac{ft}{s}-\frac{ft}{s^2} = \frac{ft \cdot s-ft}{s^2}=\frac{ft(s-1)}{s^2}=what just happened????[/math]
To extract the part of the problem without bothering with the entire context I am looking at, its saying that
[math](16.1 ft/s^2)t^2=(40 ft/s)t-(40 ft/s)+(16.1 ft/s^2)t^2-(32.2ft/s^2)t+(16.1ft/s^2)[/math][math]7.8t = 23.9[/math][math]t=23.9/7.8[/math][math]t=3.06s[/math]
What is happening when you subtract ft/s^2 from ft/s to result in 7.8t=23.9? Thats all I really have a problem with. If I know that I can try to understand the rest of the problem.
I understand arithmetic can be done with units so I have a superficial understanding of something like:
[math]\frac{(m/s)^2}{m/s^2}=\frac{m^2/s^2}{m/s^2}=\frac{m^2}{s^2} \div \frac{m}{s^2}=\frac{m^2}{s^2} \times \frac{s^2}{m}=m[/math]
but if I were to do the same according to my example I refer to, assuming its not wrong, I'd have to try something like this:
[math]\frac{ft}{s}-\frac{ft}{s^2} = \frac{ft \cdot s-ft}{s^2}=\frac{ft(s-1)}{s^2}=what just happened????[/math]