Kinematics of a Particle Tangent-Normal components

[teknodude]

Homework Statement

http://img136.imageshack.us/img136/1930/carrs8.th.png

Homework Equations

a = (v^2/p) en + (v*)et
a- accel
v-velocity
en- inward normal
et- unit vector along tangent
v*- tangential acceleration
p- curvature

The Attempt at a Solution

FOr the first part I just need to find the velocity using the above equation. Since velocity is constant a =0. Therefore tangential accel (at) = 0.4g.
Then just solve for v. Now they gave us the dimensions of the car, so I guess I can't treat the car as a just a particle. The TA just told us to subtract the curvature from the given dimensions,BUT why are we doing that?

so curvature = 120m - 0.6

 

[doodle]

I believe the adjustment is made because the center of gravity of the car is slightly above the road surface.

 

[piercas]

m = 119.4m I would like to clarify a few things about the provided homework statement and equations. Firstly, the term "particle" in kinematics refers to a point mass, which means that its size and shape are negligible compared to the distance it travels. Therefore, it is appropriate to treat the car as a particle in this problem.

Secondly, the equation provided for acceleration, a = (v^2/p) en + (v*)et, is known as the tangential-normal component form of acceleration. This means that the acceleration of the particle can be broken down into two components: the tangential component (v*)et and the normal component (v^2/p) en. The tangential component represents the change in speed of the particle along its path, while the normal component represents the change in direction of the particle's velocity.

Thirdly, the statement "since velocity is constant a = 0" is not entirely accurate. While it is true that the magnitude of the velocity is constant, the direction of the velocity changes due to the normal component of acceleration. Therefore, the particle is not moving at a constant speed, but rather at a constant velocity (constant speed in a specific direction).

Now, let's address the solution attempt. The tangential acceleration (v*)et is given as 0.4g, which means that the speed of the particle is increasing at a rate of 0.4g m/s^2. This does not mean that the tangential acceleration is equal to 0.4g. The tangential acceleration is actually equal to the rate of change of the speed of the particle, which is 0.4g m/s^2 in this case.

Next, the equation provided for acceleration is not applicable in this problem since it is for a curved path. The given dimensions of the car do not represent a curved path, but rather a straight path. Therefore, the curvature cannot be subtracted from the dimensions of the car.

In conclusion, to solve this problem, the following steps can be taken:

1. Identify the given information and what needs to be found. In this case, the given information is the tangential acceleration (0.4g m/s^2) and the dimensions of the car (120 m and 0.6 m), and what needs to be found is the velocity of the car.

2. Use the equation for tangential acceleration, a