Velocity and acceleration.

[fahd]

velocity & acceleration PLZZZZ HELP ME!

hi i have this question from fowles anlaytical mechanics.It says

A racing car moves along a circular track of radius 'a'.The speed of the car varies with time as v=kt where k is a poistive constant.Show that the angle between the velocity vector and acceleration vector is 45 degrees when t= (a/k)^1/2

i took the equation of motion to be r= a cos[(kt^2)/a]+a sin[(kt^2)/a]
then differentiated to get velocity and acceleration...i represented a particular portion of each velocity and acceleration as a unit vector p and q so as to reduce the size of the equations.HOwever when i try to calculate the angle between a and v..at that time..it doesn't come out to be 45..Can ne one please help ,me!

 

[Tide]

You can determine both the tangential and radial (centripetal) components of velocity and acceleration at any instant. Make it easy on yourself and choose your coordinates for a given instant of time to be such that the tangential direction is, say, in the postitive y direction and the radial component in the x direction.

Then note that $\vec v \cdot \vec a = v a \cos \theta$ and you can explicitly evaluate the dot product since you know the components of each vector. What must t be in order for $\theta$ to be 45 degrees?

 

[quicknote]

I am happy to help you with this question. First, let's define velocity and acceleration. Velocity is the rate of change of an object's position with respect to time, while acceleration is the rate of change of an object's velocity with respect to time. In this case, the velocity of the racing car is given by v = kt, where k is a positive constant. This means that the car's velocity is directly proportional to time.

Now, let's consider the position of the car at a particular time t. Using the equation of motion provided, we can represent the car's position as r = a cos[(kt^2)/a] + a sin[(kt^2)/a]. When we differentiate this equation with respect to time, we get the velocity as v = -2a sin[(kt^2)/a] + 2a cos[(kt^2)/a]. Similarly, when we differentiate the velocity equation, we get the acceleration as a = -4a^2 cos[(kt^2)/a] - 4a^2 sin[(kt^2)/a].

Now, let's consider the angle between the velocity and acceleration vectors at time t. We can represent the velocity vector as v = p v0, where p is a unit vector and v0 is the magnitude of the velocity. Similarly, we can represent the acceleration vector as a = q a0, where q is a unit vector and a0 is the magnitude of the acceleration.

To calculate the angle between the two vectors, we can use the dot product formula as follows: v · a = |v| |a| cos θ, where θ is the angle between the two vectors. Substituting the expressions for velocity and acceleration, we get p v0 · q a0 = |p v0| |q a0| cos θ. Simplifying this, we get cos θ = (p · q) (v0/a0).

Now, let's substitute the expressions for p and q. Since these are unit vectors, p · q = cos θ, where θ is the angle between the two vectors. Substituting this in the previous equation, we get cos θ = (cos θ) (v0/a0). Simplifying this, we get cos θ = v0/a0.

Since we know that v0 = kt and a0 = k