# Homework H1.C - Sp24

DISCUSSION:

As you know, the velocity of P is tangent to the path (where the path here is the coaster track), whereas the acceleration has components both tangent and normal to the path:

v = vet
a = v_dot et + (v2/rho) en

You are provided equations for v, rho and theta in terms of the distance s traveled along the path. Therefore, you can simply substitute those expressions into the above equations for the velocity and acceleration vectors. The only remaining task is to find the rate of change of speed, v_dot. Since the speed v is given not as a function of time, but instead as a function of s, we need to use the chain rule of derivatives to find v_dot; that is:

v_dot = dv/dt = (dv/ds)*(ds/dt) = v*(dv/ds)

# Homework H1.D - Sp24

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*et
a = v_dot*et + (v2/ρ)*en

# Homework H1.A - Sp24

 Problem statement Solution video https://youtu.be/XxGXK9Xdwqc

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:

vP = x_dot i + y_dot j
aP =
x_ddot i + y_ddot j

# Homework H1.B - Sp24

 Problem statement Solution video https://youtu.be/WshBY9Bj8jU

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.

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.