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

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The discussion centers on the motion of a particle P that starts and returns to point O after 100 seconds, with its velocity defined by the equation v(t) = 0.0001t^3 - 0.015t^2 + 0.5t. Key findings include that the particle's acceleration is zero at t = 21.1s (v = 4.81 m/s) and t = 78.9s (v = -4.81 m/s). The maximum distance from point O occurs at t = 50s, resulting in a distance of 156.25 meters. The participants clarify the velocity function and its implications for the particle's motion.

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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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