Dynamics help: variable acceleration and angular velocity

[medynamics]

I have to do these 2 problems for homework, but I can't figure them out. Any help would be appreciated.

1. The motorcycle is traveling at 1m/s when it is at A(the origin). If the speed is then increased at $\dot{V} = .1 m/s^2$, determine its speed and acceleration at the instant t = 5 seconds. The motorcycle is following the curve $y = 0.5 x^2$.2. Rod OA rotates counterclockwise with a constant angular velocity of θ = 6 rad/s. Through mechanical means collar B moves along the rod with a speed of $r = 4 t^2 m/s$, where t is in seconds. If r = 0 when t = 0, determine the magnitudes of velocity and acceleration of the collar when t = 0.75 sec.

This is the link to the image for question 2:
http://s1308.beta.photobucket.com/user/jackcastle618/media/Dynamics4_zps5a431c20.jpg.html

 

[CaptainEvil]

For question 1, start with the Kinematic Equations, specifically v = u + at

 

[medynamics]

so using v = u + at:

v = 1m/s + .1(5)
v = 1.5 m/s

now I need a. i believe i need to find radial and tangential a?
so i have a = vdot
so a_r = .1 m/s^2
but how do I get the acceleration that is based on the curve y = .5x^2?

 

[haruspex]

For question 1, start with the Kinematic Equations, specifically v = u + at

I don't think you can use that here. The rate of increase of speed is constant, but that's not the same as acceleration as strictly defined. Acceleration is a vector, or a component of a vector in a consistent direction.
medynamics, try to write down some equations relating dx/dt, dy/dt etc.

It's not a good idea to put two unrelated problems in the same thread.

 

[Nickie]

1. To solve this problem, we need to use the equations of motion for variable acceleration. We know that the initial speed (u) is 1m/s and the acceleration (\dot{V}) is 0.1 m/s^2. Using the equation v = u + at, we can find the final speed (v) at t = 5 seconds:

v = 1 + (0.1)(5) = 1.5 m/s

To find the acceleration, we can use the equation a = \dot{V}. Therefore, the acceleration at t = 5 seconds is 0.1 m/s^2.

Now, to find the speed and acceleration at any point on the curve y = 0.5 x^2, we need to use the equations of motion for variable acceleration in two dimensions. The equations are:

v_x = u_x + a_x t
v_y = u_y + a_y t
x = u_x t + 0.5 a_x t^2
y = u_y t + 0.5 a_y t^2

Since the motorcycle is following the curve y = 0.5 x^2, we can substitute this equation into the equations above and solve for v_x and v_y:

v_x = u_x + a_x t = 0 + 0.1 t = 0.1 t
v_y = u_y + a_y t = 0 + 0.5 t = 0.5 t

Therefore, the speed at t = 5 seconds is:

v = √(v_x^2 + v_y^2) = √(0.1^2 + 0.5^2) = 0.51 m/s

And the acceleration at t = 5 seconds is:

a = √(a_x^2 + a_y^2) = √(0.1^2 + 0.5^2) = 0.51 m/s^2

2. To solve this problem, we need to use the equations of rotational motion. We know that the angular velocity (θ) is constant at 6 rad/s and the speed (r) is given by r = 4 t^2 m/s. Therefore, to find the velocity and acceleration at t = 0.75 seconds, we can use the following equations:

v = rθ