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Isaac physics problem

I'm struggling to get the correct answer on the attached problem, the answer I keep getting is 4.4 (2s.f) for which it tells me the sign at some point during my calculation is incorrect even if I try to reverse engineer the problem (use different values of s till I get the correct value for t) I still get 4.4. I'm not sure if my understanding of the question is wrong or my calculations are wrong. If it would help I can post my answers to the previous parts.
(part D)
https://isaacphysics.org/questions/running_return

(edited 4 years ago)

Original post by HClarkson

I'm struggling to get the correct answer on the attached problem, the answer I keep getting is 4.4 (2s.f) for which it tells me the sign at some point during my calculation is incorrect even if I try to reverse engineer the problem (use different values of s till I get the correct value for t) I still get 4.4. I'm not sure if my understanding of the question is wrong or my calculations are wrong. If it would help I can post my answers to the previous parts.

https://isaacphysics.org/questions/running_return

Which part of the problem do you have issue? Be specific, please.
Post the working to the part of the question that you get it wrong.

Reply 2

hennersfl

OP

15

Original post by Eimmanuel

Which part of the problem do you have issue? Be specific, please.
Post the working to the part of the question that you get it wrong.

Apologies I thought I had. It's part C that I'm struggling with.

One of a few ways that I have got to 4.4 is by using s=.5*a*t^2 (as u = 0, or that's the way I understood the question) a is given as 6.9 and t can be calculated from a bit of Pythagoras and suvat to be around 1.1322 (depending on the accuracy which may be a source of error for me) putting it in the equation got me 4.4 (2 s.f) which is wrong and gets the message mentioned above.

Original post by HClarkson

I'm struggling to get the correct answer on the attached problem, the answer I keep getting is 4.4 (2s.f) for which it tells me the sign at some point during my calculation is incorrect even if I try to reverse engineer the problem (use different values of s till I get the correct value for t) I still get 4.4. I'm not sure if my understanding of the question is wrong or my calculations are wrong. If it would help I can post my answers to the previous parts.

https://isaacphysics.org/questions/running_return

Original post by HClarkson

Apologies I thought I had. It's part C that I'm struggling with.

One of a few ways that I have got to 4.4 is by using s=.5*a*t^2 (as u = 0, or that's the way I understood the question) a is given as 6.9 and t can be calculated from a bit of Pythagoras and suvat to be around 1.1322 (depending on the accuracy which may be a source of error for me) putting it in the equation got me 4.4 (2 s.f) which is wrong and gets the message mentioned above.

It seems that you have interpreted the part C of the question wrongly.

Let the time taken for the player to move from B to C be Δt.

The player speeds up uniformly (constant acceleration with magnitude a) for a duration of 0.5Δt and then slows down uniformly (constant acceleration with magnitude a) for the next 0.5Δt.

The Δt is NOT 1.1322 s NOT because of the precision

Reply 4

hennersfl

OP

15

Original post by Eimmanuel

It seems that you have interpreted the part C of the question wrongly.

Let the time taken for the player to move from B to C be Δt.

The player speeds up uniformly (constant acceleration with magnitude a) for a duration of 0.5Δt and then slows down uniformly (constant acceleration with magnitude a) for the next 0.5Δt.

The Δt is NOT 1.1322 s NOT because of the precision

I said the wrong question sorry. I meant to say part D.

(edited 4 years ago)

Original post by HClarkson

I said the wrong question sorry. I meant to say part D.

Okay.

Original post by HClarkson

Apologies I thought I had. It's part C that I'm struggling with.

One of a few ways that I have got to 4.4 is by using s=.5*a*t^2 (as u = 0, or that's the way I understood the question) a is given as 6.9 and t can be calculated from a bit of Pythagoras and suvat to be around 1.1322 (depending on the accuracy which may be a source of error for me) putting it in the equation got me 4.4 (2 s.f) which is wrong and gets the message mentioned above.

If it is part D, this description makes senses. Actually, your working makes no sense because the player needs to reach zero velocity before moving in the opposite direction.

The above graph (velocity versus time) describes the problem and what you need to solve for the problem. At time t, the player changes the direction of acceleration. For the next Δt (which was what you calculated, Δt = 1.132 s) duration, the player is trying to move back to B.

Reply 6

hennersfl

OP

15

Ah so that Δt is the time the player has to deaccelerate to zero and accelerate to B. I think I got confused with the instantaneous part.
Thanks

Original post by HClarkson

Ah so that Δt is the time the player has to deaccelerate to zero and accelerate to B. I think I got confused with the instantaneous part.
Thanks

You can say so. I would not suggest using words like decelerate or deceleration but as long as you know what you are describing, I think it is good.

I find part D too and need to read the question a few times before knowing what the question is asking. IMO the confusing phrase is the “maximum distance”… :smile:

Reply 8

hennersfl

OP

15

Original post by Eimmanuel

You can say so. I would not suggest using words like decelerate or deceleration but as long as you know what you are describing, I think it is good.

I find part D too and need to read the question a few times before knowing what the question is asking. IMO the confusing phrase is the “maximum distance”… :smile:

Ok, thank you. Would you be able to give any more pointers on how to get to the right answer as I still seem to be making mistakes?

Original post by HClarkson

Ok, thank you. Would you be able to give any more pointers on how to get to the right answer as I still seem to be making mistakes?

Okay. See the graph below where I add some labels of “points” for me to give hints

The time taken to go from O to M is t, what can you say about the time taken to go from M to N in terms of t? Think in terms of symmetry.

Displacement is the area under the velocity-time graph.
Area of OMN is equal to the area under NQ. Why?

This should help you to solve the problem.

Reply 10

hennersfl

OP

15

Original post by Eimmanuel

Okay. See the graph below where I add some labels of “points” for me to give hints

The time taken to go from O to M is t, what can you say about the time taken to go from M to N in terms of t? Think in terms of symmetry.

Displacement is the area under the velocity-time graph.
Area of OMN is equal to the area under NQ. Why?

This should help you to solve the problem.

I feel like I understand what you're saying (and you're really good at explaining this so thank you) but I keep getting the wrong answer.

The calculation I've done now is .5*6.9*(Δt/4)^2

I'm sorry that I'm being so difficult.

Original post by HClarkson

I feel like I understand what you're saying (and you're really good at explaining this so thank you) but I keep getting the wrong answer.

The calculation I've done now is .5*6.9*(Δt/4)^2

I'm sorry that I'm being so difficult.

Not sure how you get 5*6.9*(Δt/4)^2. If you want anyone to explain what is wrong, write what you have done.

You need to set up an equation in terms of t (which is the unknown) using the info: Area of OMN is equal to the area under NQ.

Reply 12

hennersfl

OP

15

Original post by Eimmanuel

Not sure how you get 5*6.9*(Δt/4)^2. If you want anyone to explain what is wrong, write what you have done.

You need to set up an equation in terms of t (which is the unknown) using the info: Area of OMN is equal to the area under NQ.

Should I get a quadratic in terms of t?

Original post by HClarkson

Should I get a quadratic in terms of t?

Yes

Reply 14

hennersfl

OP

15

Original post by Eimmanuel

Yes

And the value/s of t that I get can be used in s= ut + .5at^2, where u = 0 so s = .5*at^2 ?

Original post by HClarkson

And the value/s of t that I get can be used in s= ut + .5at^2, where u = 0 so s = .5*at^2 ?

Yes and no. You have to think what is the meaning of the displacement that you obtain from s= 0.5*a*t^2.
Physics is NOT plug in the value and calculate. You need to think about what the meaning of the answer.