Question about path resulting from acceleration

  • #1
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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.
 

Answers and Replies

  • #2
CWatters
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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?
 
  • #3
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x: 0.5 * acceleration in x direction * t^2
y: 0.5 * acceleration in y direction * t^2
 
  • #4
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.
 
  • #5
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@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!
 
  • #6
CWatters
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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.
 
  • #7
TSny
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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?
 
  • #8
TSny
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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.
 
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  • #9
CWatters
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Perhaps the RoboNerd can confirm that the problem statement is exactly as written in the OP?

It does look like there is something missing.
 
  • #10
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.
 

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