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

In summary: You can sketch the graph from t=0 to t=100, right?ok, I’ll buy that function. You can sketch the graph from t=0 to t=100, right?Yes, I can sketch the graph from t=0 to t=100.
  • #1
Shah 72
MHB
274
0
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.
 
Mathematics news on Phys.org
  • #2
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 \(\displaystyle v = 0.003t^3 - 0.015t^2 + 0.5t\)

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

-Dan
 
  • #3
topsquark said:
Your problem statement has a typo. I used your answer to 1) to get that \(\displaystyle v = 0.003t^3 - 0.015t^2 + 0.5t\)

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.
 
  • #4
topsquark said:
Your problem statement has a typo. I used your answer to 1) to get that \(\displaystyle v = 0.003t^3 - 0.015t^2 + 0.5t\)

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!
 
  • #5
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 \(\displaystyle v = 0.003t^3 - 0.015t^2 + 0.5t\)

$v(t) \ge 0$ 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?
 
  • #6
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
 
  • #7
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.
 
  • #8
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
 
  • #9
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.
 

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

1. What is motion in a straight line?

Motion in a straight line refers to the movement of an object along a single, straight path. This type of motion is also known as linear motion.

2. What is the difference between speed and velocity?

Speed is a measure of how fast an object is moving, while velocity is a measure of both speed and direction of motion. In other words, velocity takes into account the direction of an object's motion, while speed does not.

3. How is acceleration related to motion in a straight line?

Acceleration is the rate of change of an object's velocity over time. In motion in a straight line, acceleration can be positive (speeding up) or negative (slowing down) depending on the direction of the object's motion.

4. What is the equation for calculating average speed?

The equation for average speed is: speed = distance/time. This means that to calculate the average speed of an object, you divide the distance it has traveled by the time it took to travel that distance.

5. How does the slope of a distance-time graph relate to an object's speed?

The slope of a distance-time graph represents the object's speed. A steeper slope indicates a higher speed, while a flatter slope indicates a lower speed. A horizontal line on the graph means the object is not moving (zero speed).

Similar threads

Replies
4
Views
1K
Replies
2
Views
1K
Replies
6
Views
897
Replies
4
Views
2K
Replies
98
Views
5K
Replies
12
Views
641
Replies
25
Views
2K
Back
Top