Ask and answer questions here. You learn both ways.

DISCUSSION and HINTS

As expected, the acceleration of P has both non-zero tangential and normal components.

• From the equation provided for speed as a function of distance traveled, the speed of P is monotonically decreasing over the range of motion shown in the animation below. Therefore, the tangential component of acceleration always points "backward" of the direction of motion.
• The normal component decreases as P moves along the track since the speed of P is decreasing.

Can you see these two things in the animation below?

Recall that the general path description velocity and acceleration equations are given by the following:

v = v*e_t
a = v_dot*e_t + (v²/ρ)*e_n

Ask and answer questions here. You learn both ways.

Discussion

Consider the animation below of the motion for particle P:

• For all time, P moves with a positive x-component and a negative y-component of both velocity and acceleration. Why? Can you see this in the animation?
• As we will see in the next class period when we talk about the "path description" of kinematics, the velocity of a point is always tangent to the path of the point, and the acceleration always points "inward" on the path of the point. Do you see this in the animation?

The components of velocity and acceleration of P are found directly from the time derivatives of the position components:

v_P = x_dot i + y_dot j
a_P = x_ddot i + y_ddot j

Given: A particle P travels on a path described by the Cartesian coordinates of y = cx(b-x). The x-component of velocity for P is a constant.

Ask and answer questions through the "Leave a Comment" link above. You can learn as much by answering the questions of others as you do from asking questions.

From the animation below, you see that the velocity of P is always tangent to the path, and that the acceleration always points inward of the path. In the next class when we discuss the path description of kinematics, we will discover that this is true in general.

Note below that in this case the acceleration of P is always vertical. Do you know why?

HINTS:
The path of P is not given in terms of x- and y-coordinates that are known explicit functions of time (instead, the y-coordinate depends explicitly on the x-coordinate). For velocity and acceleration, we need the derivatives of x and y with respect to time. How do we do this? Recall the chain rule of differentiation:

dy/dt = (dy/dx)*(dx/dt)

Use this in setting up the components of velocity and acceleration.