# Isaac physics problem

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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

(part D)

https://isaacphysics.org/questions/running_return

Last edited by hennersfl; 1 year ago

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#2

(Original post by

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

**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

Post the working to the part of the question that you get it wrong.

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(Original post by

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.

**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.

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.

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#4

**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

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.

**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 Δ

*t*.

The player speeds up uniformly (constant acceleration with magnitude

*a*) for a duration of 0.5Δ

*t*and then slows down uniformly (constant acceleration with magnitude

*a*) for the next 0.5Δ

*t*.

The Δ

*t*is NOT 1.1322 s NOT because of the precision

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(Original post by

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 Δ

The player speeds up uniformly (constant acceleration with magnitude

The Δ

**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 Δ

*t*.The player speeds up uniformly (constant acceleration with magnitude

*a*) for a duration of 0.5Δ*t*and then slows down uniformly (constant acceleration with magnitude*a*) for the next 0.5Δ*t*.The Δ

*t*is NOT 1.1322 s NOT because of the precision
Last edited by hennersfl; 1 year ago

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#6

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

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

**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.

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 Δ

*t*(which was what you calculated, Δ

*t*= 1.132 s) duration, the player is trying to move back to B.

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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

Thanks

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#8

(Original post by

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

**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

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”…

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(Original post by

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”…

**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”…

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#10

(Original post by

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?

**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.

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(Original post by

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

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.

**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.

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

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

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#12

(Original post by

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.

**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.

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.

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(Original post by

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.

**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.

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(Original post by

Yes

**Eimmanuel**)Yes

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#16

(Original post by

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 ?

**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 ?

Physics is NOT plug in the value and calculate. You need to think about what the meaning of the answer.

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