When Does the Particle Reach Its Maximum Positive X Position?

In summary, the position of a particle moving along the x-axis depends on time as x(t) = 100 t^2 - 8 t^3 where x is in meters and t in seconds. At what time does the particle reach its maximum positive x position? The length of the path covered by the particle between t = 0.0000 s and t = 10.833 s is 9.6 meters. What is the displacement of the particle between t = 0.0000 s and t = 10.8333 s? The displacement is 0.
  • #1
bmadkins
1
0

Homework Statement


The position of a particle moving along the x-axis depends on time as x(t) = 100 t^2 - 8 t^3 where x is in m and t in s. At what time does the particle reach its maximum positive x position?

What is the length of the path covered by the particle between t = 0.0000 s and t = 10.833 s?

What is the displacement of the particle between t = 0.0000 s and t = 10.8333 s?


Homework Equations


x(t) = 100 t^2 - 8 t^3



The Attempt at a Solution


I haven't had any attempts that have led me anywhere, really. This is the first time I've encountered this type of problem.
 
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  • #2
The length of the path would be the arc length between the two points.

The displacement would be the straight line distance between the two points.
 
  • #3
Hello bmadkins,

Welcome to Physics Forums!
bmadkins said:
I haven't had any attempts that have led me anywhere, really. This is the first time I've encountered this type of problem.
The best way to do the first part of problem (finding the time at which the position is positive maximum) is to open up your Calculus I textbook or notes and review the section of how to find the local maximums and local minimums of a function.

But perhaps a more intuitive guide (which is the same thing applied to this problem), is to consider that at some moment in time, the particle is moving away from its original position. Eventually, it slows down, momentarily stops, and then goes back in the other direction toward the original position.

So at what point was it farthest away from the original position? At the instant it momentarily stopped -- at the instant it "turned around." At that point its instantaneous velocity is zero. So part of this problem is figuring out at what time (or times) the particle's velocity is zero.

So to start, you might want to ask yourself, "what is a particle's velocity (as a function of time) if you already know what its position is (also as a function of time)?"

(And I don't mean s = vt, since the particle is not moving at a constant velocity. And I don't mean v = at or s = ½at2 either, since the acceleration is not necessarily uniform. There is a more fundamental relationship between position and velocity.)
 

What is the definition of displacement?

Displacement refers to the distance and direction of an object's change in position from its starting point. It is a vector quantity, meaning it has both magnitude and direction.

What is the equation for calculating displacement?

The equation for calculating displacement is Δx = xf - xi, where Δx represents the change in position, xf is the final position, and xi is the initial position.

How is displacement different from distance?

Displacement takes into account the direction of an object's movement, while distance only measures the total length traveled without considering direction. Displacement can be equal to or less than the distance traveled.

What is a displacement vs time graph?

A displacement vs time graph is a visual representation of an object's displacement over a period of time. The x-axis represents time and the y-axis represents displacement. The slope of the graph represents the object's velocity.

How can displacement vs time problems be solved?

Displacement vs time problems can be solved by using the equation Δx = vavg x t, where vavg is the average velocity and t is the time interval. It is important to pay attention to the direction of the displacement, as it can affect the sign of the answer.

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