# Particle at rest

## Homework Statement

A particle's position is r =(ct^2−2dt^3) i hat+(2ct^2−dt^3)j hat. where c and d are positive constants.
part a) Find expressions for times t > 0 when the particle is moving in the x-direction.
part b) Find expressions for times t > 0 when the particle is moving in the y-direction.

Is there any time,t >0 when the particle is c) at rest and d) accelerating in the x-direction?
If either answer us "yes", find the time(s).

## The Attempt at a Solution

a)

r''(t)→ = d^2/dt^2 [ ct^2 -2dt^3] i
a(t)→ = 2c - 12dt
t = (2c - a)/12d

b)

r"(t)→d^2/dt^2[ 2ct^2 - dt^3] j
a(t)→ = (4c -6dt)j
t = (4c - a)/6d

c)

If particle is at rest, then velocity = 0

v→ = r'(t)→ = (2ct - 6dt^2) i + (4ct - 3dt^2) j
0 = (2ct - 6dt^2) i + (4ct - 3dt^2) j

how do I proceed from here?

Related Introductory Physics Homework Help News on Phys.org
For (a) and (b), what does "moving in x or y direction" mean?

For (c), find t such that the equation is satisfied.

For (a) and (b), what does "moving in x or y direction" mean?

For (c), find t such that the equation is satisfied.
I suppose it means (x,0) and (0,y). Word for word from the book.
Part (a) and (b) is correct. How should I attempt part(c)?
I set dr/dt to be zero but I can't reduce the expression.

How do you know that (a) and (b) are correct? I have a very different interpretation of those parts.

For (c), you have a vector equation. That means you have equations for each component. Find a common solution for the component equations, if it exists.

How do you know that (a) and (b) are correct? I have a very different interpretation of those parts.

It seems to make sense, although, I would rule out my answer being a probable falsity.
Let's begin from part(a) then.

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When you hear "something is moving in some direction", is this a statement about the position of something, or its velocity, or its acceleration?

When you hear "something is moving in some direction", is this a statement about the position of something, or its velocity, or its acceleration?
Position. But this change in position could further imply the velocity is changing, and, therefore, a change in the acceleration too.

Position. But this change in position could further imply the velocity is changing, and, therefore, a change in the acceleration too.
So if you are at x = 5 m and y = 6 m, which direction are you moving?

So if you are at x = 5 m and y = 6 m, which direction are you moving?
50° from the origin-origin being (0,0)

Edit: It cannot be ascertain.

Even if you are standing still at that position?

Even if you are standing still at that position?
I edited the post you responded to.
It cannot be ascertain unless the change in position is made relative to another position, otherwise, the question becomes meaningless. Is that what you were trying to expound?

I edited the post you responded to.
It cannot be ascertain unless the change in position is made relative to another position, otherwise, the question becomes meaningless. Is that what you were trying to expound?
Yes. And how do you characterize "change in position"?

Yes. And how do you characterize "change in position"?
Δs = vit + 0.5at^2

That is valid only for uniformly accelerated motion. Imagine ##\Delta t## is very very small. How can you approximate ##\Delta s##?

That is valid only for uniformly accelerated motion. Imagine ##\Delta t## is very very small. How can you approximate ##\Delta s##?
change in position/ change in time = velocity.

v(t) . Δt = Δs = sf - si

or in other words; the derivative of position x change in time

How do you know that (a) and (b) are correct? I have a very different interpretation of those parts.

For (c), you have a vector equation. That means you have equations for each component. Find a common solution for the component equations, if it exists.
Here's another go:

a)

If we are concern only with the velocity of the particle along the x-axis, then, this implies that the velocity along the y-axis is hypothetically = 0ms^-1

v = (vx=0,vy)
vy = 0ms^-1
vy = (dr/dt)y = 0ms^-1
ry→ = (2ct^2 -dt^3)j
dr/dt = 4ct - 3dt^2
4ct - 3dt^2 = 0
4c = 3dt
t = 4c/3d

b)

v = (vx,vy=0)
vx = 0ms^-1
vx = (dr/dt)x = 0ms^-1
rx→ = (ct^2 -2dt^3)i
(dr/dt)x = (2ct -6dt^2)i
2ct = 6dt^2
2c = 6dt
t = 2c/6d =c/3d

c)

if the particle is at rest, that implies that v = (vx = 0, vy = 0)

at t = 4c/3d, vy = 0
at t = c/3d, vx = 0

there is no time at which tx = ty, so, the particle is never at rest.

d)

particle accelerating only in the x-axis implies it's acceleration along the y-axis = 0

py'(t)→ = 4ct - 3dt^2
py"(t)→ = a(t)→ = 4c - 6dt

4c = 6dt
t = 2c/3d

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Your method and results look good to me.

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