Addition or subtraction of of x ft/s and y ft/s^2

[Adrian_B]

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]

 

[Deleted member 4993]

(16.1ft/s2)t²=(40ft/s)t−(40ft/s)+(16.1ft/s2)t²−(32.2ft/s2)t+(16.1ft/s²)

Those two terms are dimensionally incorrect. Please check your source or calculation

 

[Harry_the_cat]

To add or subtract, the units must be the same. You can't make sense of adding an acceleration with a velocity.

 

[Harry_the_cat]

Those two terms are dimensionally incorrect. Please check your source or calculation

The second last term is too, isn't it?

 

[Deleted member 4993]

The second last term is too, isn't it?

Yes indeed .... that one escaped my sleepy stare....

 

[Dr.Peterson]

To extract the part of the problem without bothering with the entire context I am looking at, its saying that

Please include the entire context. Without it, we can't tell where the error is, or whether what you show means something other than what it says.

 

[blamocur]

All your [imath]ft/s[/imath] are multiplied by [imath]t[/imath], but all [imath]ft/s^2[/imath] are multiplied by [imath]t^2[/imath], which means the resulting dimensions ([imath]ft[/imath]) are the same across your equation.

 

[Dr.Peterson]

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]

To restate my request for context:

The work is fine, solving the equation [math]16.1t^2=40t-40+16.1t^2-32.2t+16.1[/math]
The only issue is whether the units in the equation make sense; according to them you are adding distances in feet and speed in feet per second. In order to help, we have to see where the equation came from, not what is done to solve it.

If you inserted the dimensions yourself, then we need to see why you did so, and correct your thinking. If they are part of a solution given in a textbook, we need to see why they did that. In either case, that means seeing the problem and how the equation was derived from it.

 

[Adrian_B]

Thanks to all for caring.
Original question:

One ball is dropped from a cliff. A second ball is thrown down 1.00 s later with an initial speed of 40.0 ft/s. How long after the second ball is thrown will the second ball overtake the first?

Entire example solution:

Initial speed of the first ball [math]V_{1i}=0[/math]Second ball is thrown down 1.00s later with initial speed of 40ft/s [math]V_{2i}=40ft/s[/math]Distance covered by first ball in [math]t[/math] seconds will cross the second ball in [math](t-1)s[/math]
The distance covered by the first ball in t seconds is[math]x_{1}=v_{1i}t+at^2\\ =0+1/2(32.2 ft/s^2)t^2\\=(16.1 ft/s^2)t^2 .......Equation 1[/math]
The distance covered by the second ball in [math](t-1)s[/math] is [math]x_{2}=v_{2i}(t-1)+1/2a(t-1)^2\\=(40ft/s)(t-1)+1/2(32.2ft/s^2)(t^2-2t+1)\\=(40ft/s)t-(40ft/s)+(16.2ft/s^2)t^2-(32.2ft/s^2)t+(16.1ft/s^2) .......Equation 2[/math]Equating 1 and 2:
[math](16.1ft/s^2)t^2 = (40ft/s)t−(40ft/s)+(16.1ft/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]Therefore the time in which the second ball will overtake the first ball [math]=t-1\\=3.06s - 1\\=2.06 s[/math]

So during this time I have been informed that what is crucially missing are the seconds units when substituting (t-1) for t. Equating Equation 1 and 2 should yield: [math](16.1ft/s^2)t^2 = (40ft/s)t−(40ft/s)(1s)+(16.1ft/s^2)t^2−(32.2ft/s^2)(t)(1s)+(16.1ft/s^2)(1s^2)\\(16.1ft/s^2)t^2=(40ft/s)t-(40ft)+(16.1ft/s^2)t^2−(32.2ft/s)(t)+(16.1ft)\\(7.8ft/s)t=23.9ft\\t=3.06 s[/math]

 

[topsquark]

They are simply applying the given information and dropping the units until the end. This is a fairly common practice. You should be able to do the unit analysis as an exercise but if your equation has the correct units to start with you can (usually) assume that you will have the correct units in the end. (Barring, of course, any unit conversions that need to be done.)

-Dan