# Question about path resulting from acceleration

## 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

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
Homework Helper
Gold Member
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?

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

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.

@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
Homework Helper
Gold Member
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
Homework Helper
Gold Member
I'm not sure I even understand the statement of the problem. I guess it means ax = ay = 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
Homework Helper
Gold Member
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.

• Sahil Kukreja
CWatters
Homework Helper
Gold Member
Perhaps the RoboNerd can confirm that the problem statement is exactly as written in the OP?

It does look like there is something missing.

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.