What is the maximum distance of P from point O during the motion?

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Discussion Overview

The discussion revolves around the motion of a particle moving in a straight line, starting and returning to a point O after 100 seconds. Participants analyze the velocity function given by v = 0.000t^3 - 0.015t^2 + 0.5t, exploring its implications for the particle's motion, including the determination of when the acceleration is zero, sketching the velocity-time graph, and finding the maximum distance from point O during the motion.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants calculate the velocity at times when acceleration is zero, identifying t = 21.1s and t = 78.9s with corresponding velocities of 4.81 m/s and -4.81 m/s.
  • There is a suggestion to sketch the velocity-time graph, indicating that it will have two curves: one increasing from 0 to 50 seconds and another decreasing from 50 to 100 seconds.
  • One participant expresses confusion about finding the greatest distance from point O during the motion.
  • Another participant points out a potential typo in the velocity function, proposing an alternative form and defining the distance function s(t) to find the maximum distance.
  • There is a discussion regarding the implications of the velocity function, with some participants arguing that if v(t) is always non-negative, the particle cannot return to point O, while others assert that the particle does return as stated in the problem.
  • Participants debate the correctness of the velocity function and its implications for the particle's motion, with one participant acknowledging the need to consider intervals where the velocity changes sign when calculating the greatest distance.

Areas of Agreement / Disagreement

Participants express disagreement regarding the interpretation of the velocity function and its implications for the particle's motion. Some assert that the particle cannot return to point O if the velocity is always positive, while others maintain that it does return as stated in the problem. The discussion remains unresolved regarding the correct interpretation of the velocity function and its effects on the motion.

Contextual Notes

There are unresolved questions about the accuracy of the velocity function and its implications for the particle's motion. The discussion includes potential typos and varying interpretations of the motion described.

Shah 72
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A particle moves on a straight line. It starts at a point O on the line and returns to O 100 s later. The velocity of P is v m/s at time t s after leaving O where v= 0.000t^3- 0.015t^2 +0.5t.
1) Find the values of v at the times for which the acceleration of P is zero
I got when t= 21.1s, V= 4.81m/s and when t= 78.9s v= -4.81m/s
2) sketch the velocity time graph for Ps motion for 0<t<100
So I plotted for t=0 v=0, t= 21.1 v=4.81, t= 78.9, v= -4.81 and t=100, v=0. So it will be two curves with the curve going up between 0 and 50 and curve going down between 50 and 100.
Is this right??
3) Find the greatest distance of P from 0 for 0<t<100
I don't understand this part.
 
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Shah 72 said:
A particle moves on a straight line. It starts at a point O on the line and returns to O 100 s later. The velocity of P is v m/s at time t s after leaving O where v= 0.000t^3- 0.015t^2 +0.5t.
1) Find the values of v at the times for which the acceleration of P is zero
I got when t= 21.1s, V= 4.81m/s and when t= 78.9s v= -4.81m/s
2) sketch the velocity time graph for Ps motion for 0<t<100
So I plotted for t=0 v=0, t= 21.1 v=4.81, t= 78.9, v= -4.81 and t=100, v=0. So it will be two curves with the curve going up between 0 and 50 and curve going down between 50 and 100.
Is this right??
3) Find the greatest distance of P from 0 for 0<t<100
I don't understand this part.
Your problem statement has a typo. I used your answer to 1) to get that [math]v = 0.003t^3 - 0.015t^2 + 0.5t[/math]

Call s(t) the distance function and define s(O) = 0. What is the greatest value of s for 0 < t < 100?

-Dan
 
topsquark said:
Your problem statement has a typo. I used your answer to 1) to get that [math]v = 0.003t^3 - 0.015t^2 + 0.5t[/math]

Call s(t) the distance function and define s(O) = 0. What is the greatest value of s for 0 < t < 100?

-Dan
So the greatest distance is when t=50s which is 156.25m.
 
topsquark said:
Your problem statement has a typo. I used your answer to 1) to get that [math]v = 0.003t^3 - 0.015t^2 + 0.5t[/math]

Call s(t) the distance function and define s(O) = 0. What is the greatest value of s for 0 < t < 100?

-Dan
Thanks very much!
 
Shah 72 said:
It starts at a point O on the line and returns to O 100 s later.

topsquark said:
Your problem statement has a typo. I used your answer to 1) to get that [math]v = 0.003t^3 - 0.015t^2 + 0.5t[/math]

$v(t) \ge 0$ for all $t \ge 0$ ... this says the particle always moves in the positive direction away from the starting point O.

How, then, can it return to point O?
 
skeeter said:
$v(t) \ge 0$ for all $t \ge 0$ ... this says the particle always moves in the positive direction away from the starting point O.

How, then, can it return to point O?
No the question says that it moves in a straight line. It starts at point O and returns to point O 100s later
 
Shah 72 said:
No the question says that it moves in a straight line. It starts at point O and returns to point O 100s later

Velocity is a vector quantity ... if v(t) > 0 for all t > 0, then the particle moves in the positive direction only. To return to its starting position, it must have a negative velocity.

Check the given velocity function again ... the one you posted will not work.
 
skeeter said:
Velocity is a vector quantity ... if v(t) > 0 for all t > 0, then the particle moves in the positive direction only. To return to its starting position, it must have a negative velocity.

Check the given velocity function again ... the one you posted will not work.
Yeah I agree you are right. The velocity between 50 to 100 is negative so it comes back. I wasn't understanding the last question which asked for the greatest distance in the given time interval between 0 and 100. Since there are more time intervals in between which is 21.1 s where the velocity is positive and 78.9s where the velocity is negative, I was thinking whether I should take these intervals too when I calculate the greatest distance
 
Shah 72 said:
The velocity between 50 to 100 is negative so it comes back …

Once again, is the velocity function $v(t) = 0.003t^3 - 0.5t^2 + 0.5t$ ?

If so, look at the graph of velocity … if not, post the correct velocity function.

59EEA5A6-7C65-4D8B-A7EC-C0F2687655C3.jpeg
 
  • #10
skeeter said:
Once again, is the velocity function $v(t) = 0.003t^3 - 0.5t^2 + 0.5t$ ?

If so, look at the graph of velocity … if not, post the correct velocity function.

View attachment 11336
V=0.0001t^3-0.015t^2+0.5t
 
  • #11
Shah 72 said:
V=0.0001t^3-0.015t^2+0.5t

ok, I’ll buy that function.
 

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