Distance , Speed , Time ? "Toda , Jim walks from a to b & back @ a steady speed of 4 mph..."

[Ronan1]

Today , Jim walks from a to b & back @ a steady speed of 4 mph . Yesterday , he walked the distances @ 3 mph & 5 mph & it took 11 minutes longer . How far is it from a to b ? Thanks in advance .

For the second walk , T = D / S is 11 greater than D / S for the first .

 

[stapel]

Today , Jim walks from a to b & back @ a steady speed of 4 mph . Yesterday , he walked the distances @ 3 mph & 5 mph & it took 11 minutes longer . How far is it from a to b ? Thanks in advance .

For the second walk , T = D / S is 11 greater than D / S for the first .

1) What is the distance "D" covered today? (Hint: Add)

2) What speed was used today? (Hint: 4mph)

3) What expression then stands for the time (in miles per hour) for his walk today? (Hint: If d = rt, then t = r/d, where "d" is "distance", "r" is "rate", and "t" is "time")

4) What distance was travelled yesterday at 3 mph? (Hint: "D")

5) What then was the time taken for this part of the walk? (Hint: 3/D)

6) What distance was travelled yesterday at 5 mph? (Hint: "D")

7) What then was the time taken for this part of the walk? (Hint: Same process as for (5).)

8) What was the total time walked yesterday? (Hint: Add)

9) How did yesterday's time relate to today's time? (Hint: +11/60)

10) Create an equation for the times. Solve for the value of D.

If you get stuck, please reply with your answers to the above, showing all of your work and reasoning. Thank you!

Eliz.

 

[The Highlander]

Today , Jim walks from a to b & back @ a steady speed of 4 mph . Yesterday , he walked the distances @ 3 mph & 5 mph & it took 11 minutes longer . How far is it from a to b ? Thanks in advance .

For the second walk , T = D / S is 11 greater than D / S for the first . No! Because speeds are in mph!
                                                            (See my 'explanation' down below.)

Alternatively,

We can see (from your post) that you know that:-

\(\displaystyle \sf{Speed=\frac{Distance}{Time}}\ \ \) ie: \(\displaystyle ~~\sf{S=\frac{D}{T}\ \implies~T=\frac{D}{S}}\)​

And in each of the walks the distance from a to b is D. (Therefore, D is constant throughout.)

So, let us call the times for each walk:-

T₄ for today’s walk at 4 mph​

T₃ for yesterday’s walk at 3 mph​

and​

T₅ for yesterday’s walk at 5 mph​

​

Then \(\displaystyle \small{\left(\textrm{since }\scriptsize{\sf{T=\frac{D}{S}}}\right)}\): \(\displaystyle ~\sf{T_4=\frac{D}{4},~ T_3=\frac{D}{3}~\textrm{and}~T_5=\frac{D}{5}}\)

Now, you know that: \(\displaystyle \sf{T_3 + T_5 = 2\times T_4 +\footnotesize{\frac{11}{60}}}\) because yesterday’s walks took 11 minutes longer than today's. NB: You have to use \(\displaystyle \frac{11}{60}\) here and not 11 because you are dealing with mph (miles per hour) and so your times must all be in hours not minutes; 11 minutes being 11 sixtieths of an hour. (This is why your comment: "For the second walk , T = D / S is 11 greater than D / S for the first" is not correct.)

But you can now replace all the times (T’s) by fractions of D.

Thus: \(\displaystyle ~\sf{T_3 + T_5 = 2T_4 +\frac{11}{60}}\)

\(\displaystyle \sf{\implies T_3+T_5-2T_4=\frac{11}{60}}\)

\(\displaystyle \sf{\implies T_3-2T_4+T_5=\frac{11}{60}}\)

\(\displaystyle \sf{\implies\small{\frac{D}{3}-2\frac{D}{4}+\frac{D}{5}}=\frac{11}{60}}\)      (Substituting the fractions of D for the corresponding times.)

\(\displaystyle \sf{\implies\small{\frac{1}{3}D-2\times\frac{1}{4}D+\frac{1}{5}D}=\frac{11}{60}}\)

\(\displaystyle \sf{\implies\small{\frac{1}{3}D-\frac{2}{4}D+\frac{1}{5}D}=\frac{11}{60}}\)

(That could be further simplified to: \(\displaystyle \sf{\small{\frac{1}{3}D-\frac{1}{2}D+\frac{1}{5}D}=\frac{11}{60}}\) but I recommend you don't do that. ?)

Can you now solve the equation: \(\displaystyle \sf{\small{\frac{1}{3}D-\frac{2}{4}D+\frac{1}{5}D}=\frac{11}{60}}\) to find D (in miles)?

(Hint: Start by replacing all the fractions by equivalent ones that all have the same denominator.)

Hope that helps. ?

 

[The Highlander]

Sheesh! I could have walked the distance from a to b there and back myself in the time it took me to get all the 'formatting' exactly how I wanted it on that! ?