LVang+-+L6+-+Motion+on+an+Incline+Lab

Lue Vang Mr. Kellogg AP Physics - Pd. 7 3 October 2011 Performed On: 28 September 2011 Lab Partners: Arielle Roth **Motion on an Incline Plane Lab** Part 1: 5.) The slope of graph 1's tangent is the velocity value of graph 2.

Part 2:



2.) The rate of change of velocity as a function of time is equal to m/s/s which is acceleration of m/s 2 . According to the data, the change in velocity is relatively constant because it lines up with the linear fit line for the freely-rolling interval. The slope's significance originates from its algebraic meaning of change in (y) over change in (x). In this case, it's the change of velocity over time. The negative symbol means velocity had to be negative, thus signifying the cart's direction. A negative slope in above the x-axis means the cart is slowing down as it moves forward, but as the line crosses the x-axis, it means the cart has changed direction and is now speed up in the direction opposite of it's original trajectory.

3.) As the elevation of the track increases, the slope of the graph increases. Also, the direction of the slope depends on how the experiment is conducted. If the radar sensor is attached to the moving vehicle, the slope is positive in the interval of free rolling. If the radar is set in a fixed position to track the cart as it moves, the slope is negative.

4.) At t = 0 is the initial velocity because the cart hasn't moved at all prior to t = 0. Hence, a general equation relating the velocity and time for an object moving with constant acceleration will be: y = mx + bv = at + v 0  v = -1.165t + 0.09029 5.) In the form of a quadratic curve fit, the equation for graph 1 will be: Position (x) = At 2 + Bt + C A = -0.7 ......... B = 1.1 x = -0.7t 2 + 1.1t + 0.4 6.) An equation relating the position and time of an object undergoing constant acceleration is as follows: Position (x) = At 2 + Bt + C  x = At 2 + Bt + x 0   x = x  0 + v 0 t + 1/2at 2