# Homework Help: Instantaneous speed

1. Aug 12, 2015

### chapp

1. The problem statement, all variables and given/known data

A 60kg patient arrives at the Reception area on a wheelchair that is 8kg. The healthcare professional has been asked to push the wheelchair to the Examination Room. The corridor from the Reception area to the Examination Room follows a circular path.

The healthcare professional pushes the wheelchair from its resting phase and maintains a steady speed of 2m/s as the wheelchair travels to its destination, which is 25m from the Reception area. It takes the Healthcare professional 80. sec to push the wheelchair patient from the Reception area to the Examination Room

2. Relevant equations

1. Define the term ‘instantaneous speed’ and deduce what its value will be when the wheelchair is 5m away

3. The attempt at a solution

Instantaneous speed of wheel chair 5m away from examination room:

Instantaneous speed= distance/time

Time (t) = 80s

Speed (s) = 2m/s

Distance (d)= sxt

= 2m/s x 80s =160m

= 160m – 5m = 155m

So, @155m( 5m away from examination room) time= ? ? Speed= ??

Time (t) of wheel chair @ 155m = 5m/160m x 100 = 3.125%
t= 3.125/80s x 100 = 3.9s
t= 80s - 3.9s = 76.094s

Speed (s) of wheel chair 5m away from examination room = d/t

s = 155m/70.094s

= 2.03m/s

Using significant digits =2.0m/s

Therefore, from the above justification, it the instantaneous speed of the wheel chair 5m away from the examination room remained at the steady speed of 2m/s. The speed of the wheel chair did not change in that ‘instant’ 5m away from the room as the patient was being pushed at a steady, unchanging speed the entire time.

Can someone please tell me if I'm on the right track or not, or give me any hints. This question has a lot of uncertainty. Not sure if i am meant to use 'waves' to figure it out?
no need to tell me the definition of instantaneous speed! thanks :D

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2. Aug 12, 2015

### billy_joule

I think the question is poorly stated, is that exactly as the book says?
I understand it that the wheelchair traverses a semicircle with diameter of 25m which is definitely not a distance of 160m as you've calculated.
That is, your assumption of instantaneous acceleration is incorrect.

3. Aug 12, 2015

### haruspex

4. Aug 12, 2015

### chapp

Yeah i thought my answer was completly wrong! The question is copied exactly word for word. I just assumed that i would state the answer is 2m/s as it is a 'Steady' speed? but then that also doesn't seem right?

5. Aug 12, 2015

### haruspex

Why semicircle? (I think chapp drew the diagram, so don't read too much into it.)

6. Aug 13, 2015

### billy_joule

I'm not so sure. Reading back now, only the diagram implies a semicircle.

Either way, a terribly worded question that doesn't appear to have a solution.
@chapp what book is this from?

7. Aug 13, 2015

### chapp

Its not out of a book, the questions were given to us from the teachers and thats the diagram that was given to us too.

8. Aug 13, 2015

### billy_joule

In that case maybe my initial interpretation was right.
That is, the distance travelled is 25m*π/2
in a time of 80s gives an average speed of ~0.5 m/s
If constant acceleration is assumed a max speed of 1 m/s is reached at the destination.

Or if we assume a constant speed of 2 m/s we end up with a travel time of ~20 seconds

I suspect the healthcare professional has travelled via the cafe and had a snack in the missing minute (or alternatively warped his perception of time after a trip to the drug cabinet).

9. Aug 13, 2015

### chapp

hahahahahah! (drug cabinet)
Thanks heaps for the help. I hope in the future the questions are more specific

10. Aug 13, 2015

### billy_joule

Let us know what the teacher says about the problem

11. Aug 13, 2015

### chapp

i defiantly will! I've never been so frustrated over a question before hahahaha

12. Aug 13, 2015

### CWatters

Are you sure it says "maintains" or is it "obtains"?

If it was "obtains" then I think you have a period of acceleration followed by a period of constant speed motion.

s = distance, v = final velocity (2ms), t = time

For the first phase..

s1 = 0.5 vt1 ................. 1

For second phase

s2 = vt2........................2

Also

s1 + s2 = 25m...............3
t1 + t2 = 80s..................4

Four variables and 4 equations so can be solved.

You can then calculate if it has reached 2m/s by the time it gets to within 5m of the end. Perhaps it hasn't, I haven't checked.

13. Aug 13, 2015

### CWatters

Another possibility is that he accelerates until he obtains 2m/s then decelerates. but then you have to take issue with the word "steady" in the problem statement.