Solution:

D = rt

Going Rate = 60 mph

Returning Rate = 40 mph

Going:

D_1 = 1(60)

D_1 = 60 miles

Returning:

D_2 = 1(40)

D_2 = 40 miles

D_1 + D_2 = answer.

Yes? No?

- Thread starter harpazo
- Start date

Solution:

D = rt

Going Rate = 60 mph

Returning Rate = 40 mph

Going:

D_1 = 1(60)

D_1 = 60 miles

Returning:

D_2 = 1(40)

D_2 = 40 miles

D_1 + D_2 = answer.

Yes? No?

\(\displaystyle d=60t=40(2-t)\implies t=\frac{4}{5}\implies d=48\text{ mi}\)

What's wrong with my reasoning?

\(\displaystyle d=60t=40(2-t)\implies t=\frac{4}{5}\implies d=48\text{ mi}\)

What's wrong with my reasoning?

...it appears you are assuming she takes 1 hour going there and 1 hour coming back.

\(\displaystyle d=vt\)

Now suppose we are given two different speeds for the same distance:

\(\displaystyle d=v_1t_1=v_2t_2\)

If we assume \(\displaystyle t_1=t_2\) then it follows that we must also have \(\displaystyle v_1=v_2\). So, when we travel a certain distance, if speed changes, then so must the time it takes to cover that distance. If we double our speed, then the time it takes to cover the distance is cut in half, etc.

Can you post and solve a word problem to show this fact?

\(\displaystyle d=vt\)

Now suppose we are given two different speeds for the same distance:

\(\displaystyle d=v_1t_1=v_2t_2\)

If we assume \(\displaystyle t_1=t_2\) then it follows that we must also have \(\displaystyle v_1=v_2\). So, when we travel a certain distance, if speed changes, then so must the time it takes to cover that distance. If we double our speed, then the time it takes to cover the distance is cut in half, etc.