D=st

[Almost done with degree]

I get confused on how to work problems like the following: A train goes 1200 miles at an average speed of 50 mph, so how long does it take to go the distance? (Solve, using the equation: d=st) So I know d (distance)=1200, s (speed)=50 mph so I then work to find t (time)=? Thats about as far as I get before I start getting all confused. Anyone have a way for me to remember what to do to find T (time)?

 

[mmm4444bot]

:idea: We may always substitute known values for their respective symbols in a given equation.

You know that the distance is 1200.

In the famous equation d = s*t, you know that symbol d represents that distance.

Hence, the value of d in your exercise is 1200, and you may write the new equation 1200 = s*t.

You know the value of s, too.

If you substitute the known value for s into the equation 1200 = s*t, then you will have an equation with only numbers and symbol t.

Solve for t in the usual way. :cool:

PS: Do not write T when you're talking about t. Stick with the given symbols. Same goes for D.

 

[Almost done with degree]

Check my answer

I came up with 24 hours to travel 1200 miles

Substitute the known values for their respective symbols in the given equation.

You know that the distance is 1200.

In the equation d = s*t, you know that symbol d represents the distance.

Hence, d = 1200, and you may write the new equation 1200 = s*t.

You know what t equals, too.

If you substitute the known value for t into the equation 1200 = s*t, then you will have an equation with only numbers and symbol s.

Solve for s in the usual way. :cool:

PS: Do not write T when you're talking about t. Stick with the given symbols. Same goes for D.

 

[mmm4444bot]

I came up with 24 hours

Oops. When I replied earlier, I was somehow thinking that speed was the unknown in this exercise. (I will fix that post.)

Despite my sloppy goof, you did good!

Distance is the product of speed*time, and 50 mph times 24 hours does indeed equal 1200 miles.

 

[soroban]

Hello, Almost done with degree!


Manipulating formulas requires only basic Algebra,
. . yet some have difficulty with the concept.

Years ago, one of my student showed me a "trick" for simple formulas.


Suppose the formula is: .\(\displaystyle D \;=\;S \times T\)

We write the formula in a "box".
Write the single letter on top, the rest on the bottom.

. . . . . \(\displaystyle \begin{array}{|c|} \hline D \\ \hline S \times T \\ \hline \end{array}\)

To solve for a letter (variable), cover that letter with your finger.
. . What remains is the formula you seek.

To solve for \(\displaystyle S\), cover \(\displaystyle S\) . . . and you see "D over T".



Note that this works for the Interest formula: .\(\displaystyle I \:=\\times R
\times T\)

. . . . . \(\displaystyle \begin{array}{|c|} \hline I \\ \hline P \times R \times T \\ \hline \end{array}\)

 

[Almost done with degree]

Thats a neat way to remember, do you have anymore tips?

This is truly a neat way to remember.