Help with my hard homework

1

14yrold

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The landscape lift travels at an angle of 15 degrees at a speed of 25 km/h. The height difference between the departure point and the elevator ending point is 2300m. How long does it take the elevator to travel? Give the answer to the nearest minute.

The only answer I got was 5 minutes, but I did not include the 15 degrees. I managed to calculate the time using only speed and distance. I have no idea how to include the 15 degrees with it.
 
Last edited:
How did you get your answer?
Do you have a picture to illustrate the problem?
 
blamocur said:
How did you get your answer?
Do you have a picture to illustrate the problem?
Click to expand...
The teacher didn't provide any pictures. He told us the problem verbally.

I got my answer by dividing the distance with speed. So 2,3/25 and I got 0.092h which is 5 minutes and 31 seconds
 
14yrold said:
The landscape lift travels at an angle of 15 degrees at a speed of 25 km/h. The height difference between the departure point and the elevator ending point is 2300m. How long does it take the elevator to travel? Give the answer to the nearest minute.

The only answer I got was 5 minutes, but I did not include the 15 degrees. I managed to calculate the time using only speed and distance. I have no idea how to include the 15 degrees with it.
Click to expand...
Since
14yrold said:
The teacher didn't provide any pictures.
Click to expand...
You would need to visualize the situation - and draw an approximate sketch. That should help you to solve the problem correctly. Can you please post a sketch as you visualize the problem? That will let us understand the "starting point" where we can help. Also label your sketch, indicating the starting point and ending point of the "lift".
 
14yrold said:
The teacher didn't provide any pictures. He told us the problem verbally.

I got my answer by dividing the distance with speed. So 2,3/25 and I got 0.092h which is 5 minutes and 31 seconds
Click to expand...

2300 is the height difference, but the lift does not travel vertically. You need to find the actual distance traveled by the lift, and a picture would be helpful.
 
To get minutes, you'll need to convert km/hr into m/min ...
 
14yrold said:
Yeah I think it's like this, but I still don't understand how to answer with minutes...
Click to expand...
Yes, you do!​

You got your 'answer' (above) in minutes with no bother at all! ?

(Although you should (very importantly) note that your answer should have been 6 minutes as you were told to give it to the nearest minute and 5min 31sec is closer to 6min than 5min!)

You don’t need to convert the speed to m/min (though there’s absolutely nothing wrong with going down that ‘route’), the method you described above (in Post #3) is perfectly valid and works fine to produce a 'correct' answer because you converted the distance (travelled) into km and recognized that this gave you an answer in hours which you then simply converted to minutes. ?

The problem, of course, is that the distance you used is not the right one to use! ?

Others have already provided sketches (like mine, below, and the one you should have tried to draw at the outset too; always start with a sketch! ?) so you should now see that the distance travelled by the ‘lift’ is not the 2,300m (it rises, vertically) but the length represented by the line SE, which is nearly 4 times the 2,300m you used, so the actual journey time will be approximately 4 times as long as the time you calculated.

rt_tri_prob.png​

So, the only ‘problem’ you have now is to work out the length, SE (since you already do know, as you have shown above, how to arrive at the journey time in minutes using just Distance & Speed).

If you are unsure how to go about that then it may help you to study the ‘Summary Sheet’ (below) that I give out to my pupils to copy into their workbooks and refer back to regularly until they know it off by heart!

The problem, as it appears to have been presented to you by your teacher, is somewhat ambiguous! Your teacher should have specified whether the lift operates at an elevation of 15° above the horizontal or 15° off the vertical (eg: if it were carrying passengers up the side of a steep cliff face; the actual distance travelled by the lift wouldn’t have been very much more than 2,300m in that case, just another 80m or so) and, had s/he done so, that might have given you a good hint as to how to ‘visualize’ the situation (and sketch it!) and you might then have realized for yourself why the 2,300m (vertical rise) given was not the distance travelled by the lift. Everyone here is assuming that the lift is travelling at a 15° incline above the horizontal and, given that it is described as a "landscape" lift, that seems like a very reasonable assumption.

However, this is another good reason why you should draw (or copy from here) a sketch and submit that with your workings & final answer. If you do so then your teacher must give you full credit even if the ‘intention’ was that the lift was travelling at 15° off the vertical (unless s/he did tell you that)!

Please come back and tell us what answer you now get for the journey time (in minutes & seconds) and to the nearest minute. We will then confirm it’s correct or offer further advice on how to adjust your working(s).

Basic Right-Angled Triangle Trigonometry Summary Sheet:-
Basic-R-A-Triangle-Trig.png
​
 
Last edited:
The Highlander said:
Yes, you do!​

You got your 'answer' (above) in minutes with no bother at all! ?

(Although you should (very importantly) note that your answer should have been 6 minutes as you were told to give it to the nearest minute and 5min 31sec is closer to 6min than 5min!)

You don’t need to convert the speed to m/min (though there’s absolutely nothing wrong with going down that ‘route’), the method you described above (in Post #3) is perfectly valid and works fine to produce a 'correct' answer because you converted the distance (travelled) into km and recognized that this gave you an answer in hours which you then simply converted to minutes. ?

The problem, of course, is that the distance you used is not the right one to use! ?

Others have already provided sketches (like mine, below, and the one you should have tried to draw at the outset too; always start with a sketch! ?) so you should now see that the distance travelled by the ‘lift’ is not the 2,300m (it rises, vertically) but the length represented by the line SE, which is nearly 4 times the 2,300m you used, so the actual journey time will be approximately 4 times as long as the time you calculated.

So, the only ‘problem’ you have now is to work out the length, SE (since you already do know, as you have shown above, how to arrive at the journey time in minutes using just Distance & Speed).

If you are unsure how to go about that then it may help you to study the ‘Summary Sheet’ (below) that I give out to my pupils to copy into their workbooks and refer back to regularly until they know it off by heart!

The problem, as it appears to have been presented to you by your teacher, is somewhat ambiguous! Your teacher should have specified whether the lift operates at an elevation of 15° above the horizontal or 15° off the vertical (eg: if it were carrying passengers up the side of a steep cliff face; the actual distance travelled by the lift wouldn’t have been very much more than 2,300m in that case, just another 80m or so) and, had s/he done so, that might have given you a good hint as to how to ‘visualize’ the situation (and sketch it!) and you might then have realized for yourself why the 2,300m (vertical rise) given was not the distance travelled by the lift. Everyone here is assuming that the lift is travelling at a 15° incline above the horizontal and, given that it is described as a "landscape" lift, that seems like a very reasonable assumption.

However, this is another good reason why you should draw (or copy from here) a sketch and submit that with your workings & final answer. If you do so then your teacher must give you full credit even if the ‘intention’ was that the lift was travelling at 15° off the vertical (unless s/he did tell you that)!

Please come back and tell us what answer you now get for the journey time (in minutes & seconds) and to the nearest minute. We will then confirm it’s correct or offer further advice on how to adjust your working(s).

Basic Right-Angled Triangle Trigonometry Summary Sheet:-
Basic-R-A-Triangle-Trig.png
​
Click to expand...
I'm not sure if this is right but
1651180890351.png1651181058769.png
Then I used the same time distance speed formula
1651181643879.png
Now I answer I got is 21 min 20 sec. I'm not even sure if this is correct at all, but I tried.
 
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