Question about path resulting from acceleration

[RoboNerd]

Homework Statement

If an object moves with non-zero, constant accelerations in two dimensions which are equal in magnitude, its path is

1) linear
2) hyperbolic
3) elliptical
4) parabolic
5) circular

Homework Equations

no equations

The Attempt at a Solution

apparently, the correct answer is 1. How is that even possible. Thanks in advance for the input.

 

[CWatters]

Take an example...

Imagine an object starts from rest at the origin x,y = 0,0
Then it starts moving and accelerates at the same rate in x and y.
After time t, what will the x and y components of the displacement be?

 

[RoboNerd]

x: 0.5 * acceleration in x direction * t^2
y: 0.5 * acceleration in y direction * t^2

 

[Sahil Kukreja]

Let Ax be the acceleration in the x direction
and Ay be the acceleration in the y direction

Let Ax=k1 ; Ay = k2

Integrate to get
Velocity
Vx = k1t + c1 ; Vy = k2t + c2

Again integrate to get
Position
Sx = (k1t^2)/2 + c1t + d1 -(I)
Sy= (k2t^2)/2 + c2t + d2 -(II)

Multiply (I) by k2 and (II) by k1
Then Subtract to get a linear relation between Sx and Sy.

 

[RoboNerd]

@Sahil Kukreja, I see how your approach works, but this question is intended to be solved in a simple manner. I believe this method overcomplicates it. Is there any way to do this with some geometrical intuition?

Thanks!

 

[CWatters]

x: 0.5 * acceleration in x direction * t^2
y: 0.5 * acceleration in y direction * t^2

acceleration in x direction = acceleration in y direction

So y = x which is a straight line.

 

[TSny]

I'm not sure I even understand the statement of the problem. I guess it means a_x = a_y = constant. But it doesn't say that the object starts at rest. It seems to me that you just have an object moving in 2 dimensions with a constant acceleration vector. Isn't the general motion parabolic, like projectile motion except the direction of the acceleration is not vertical?

 

[TSny]

Sx = (k1t^2)/2 + c1t + d1 -(I)
Sy= (k2t^2)/2 + c2t + d2 -(II)

Multiply (I) by k2 and (II) by k1
Then Subtract to get a linear relation between Sx and Sy.

Hi, Sahil.
This does not necessarily imply that the trajectory is linear because the relation still involves the time t explicitly.

Suppose y = ax + t where a is a constant and y and x are functions of time t. At time t₁, the point (x,y) will lie on the line y = ax + t₁. At time t₂, (x,y) will lie on a different line y = ax + t₂.

 

[CWatters]

Perhaps the RoboNerd can confirm that the problem statement is exactly as written in the OP?

It does look like there is something missing.

 

[Sahil Kukreja]

Hi, Sahil.
This does not necessarily imply that the trajectory is linear because the relation still involves the time t explicitly.

Suppose y = ax + t where a is a constant and y and x are functions of time t. At time t₁, the point (x,y) will lie on the line y = ax + t₁. At time t₂, (x,y) will lie on a different line y = ax + t₂.

yes i think you are correct.