0
$\begingroup$

This may seem like a silly question, so pardon in advance. I was working through some practice 2D Kinematic word problems, and I was wondering whether my approach is wrong.

I know how to calculate the acceleration in the x-direction and the velocity in the x-direction, but up until now, I've been assuming that the displacement ($x_f-x_i$) is equal to the length of the horizontal leg of the triangle formed by the ramp. Is it in fact the length of the hypotenuse? This doesn't seem right to me. The reason I'm confused is because of this practice problem. My intuition tells me that they solved this problem incorrectly.

$\endgroup$
2
$\begingroup$

The length of the hypotenuse is a displacement vector representing the sum of the displacement vector in the x direction and the displacement vector in the y direction. The distance traveled in this case is also the length of the hypotenuse, even though it is a scalar rather than a vector.

Acceleration, velocity, and displacement are vectors. Each of them is composed of a y vector and an x vector. When an object rolls down an incline, the magnitude of displacement is the length of the incline, which can be represented as the sum of the x-displacement and the y-displacement vectors, which can be computed as the hypotenuse of the right triangle they form.

$\endgroup$
  • $\begingroup$ But for an incline, there is only acceleration in the x-direction, no acceleration in the y-direction. When using the formula $V_xf^2 = V_xi^2 + 2a(x_f - x_i)$, it doesn't make sense to use the length of the hypotenuse in place of $x_f - x_i$ $\endgroup$ – AleksandrH Feb 4 '17 at 16:44
  • 1
    $\begingroup$ @AleksandrH : For an incline, there is also acceleration in the y-direction. However, the video does not make use of vector addition, and does not provide enough information to add vectors. It treats the problem as though it were 1-dimensional only. For instruction about vector addition, see: (physicsclassroom.com/class/vectors/Lesson-1/Vector-Addition) and scroll down to "Pythagorean theorem". $\endgroup$ – Ernie Feb 4 '17 at 17:10
  • $\begingroup$ From what we've learned in my course, there is no acceleration in the y-direction for an object on a ramp because the surface of the ramp "blocks" the y-component of the acceleration. Now I'm just really confused... $\endgroup$ – AleksandrH Feb 4 '17 at 19:12
  • $\begingroup$ @AleksandrH : If the x-axis were parallel to the inclined surface, there would be no acceleration in the Y-direction, which would be perpendicular to the inclined surface. Any other placement of the axes requires there to be acceleration in the y-direction. Maybe your course treats the inclined plane as though it were the x-axis? $\endgroup$ – Ernie Feb 4 '17 at 19:37
  • $\begingroup$ Oooooh, okay--reviewing the text, I noticed that they did orient the x-axis along the hypotenuse. $\endgroup$ – AleksandrH Feb 4 '17 at 21:44
1
$\begingroup$

The equations used in the video can only be used along a particular direction.In the problem that you have mentioned in your question, the acceleration has been considered to be 3.6 along the incline. In that case, the problem simply becomes 1D. All the factors are taken to be along the ramp and hence, it is not wrong.

Now, to the displacement part of your question. The displacement is indeed Δx but along the horizontal only. Vertically, it is Δy. I do not know how much of vectors you have done till now, but the the magnitude of the displacement is length of the hypotenuse i.e the square root of the sum of squares of Δx and Δy.

If you have a force that acts vertically downwards, use change in y. On the other hand, if you have force that acts on a particle horizontally, use change in x. Whichever direction the force works in, consider the change in position along that direction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.