write a discussion/conclusion for a lab report

Week of the experiment: 4

OBJECTIVE (3 points):

The objective of this lab is to gain a more complete conceptual understanding of 2-D kinematics. We will explore the concepts of position, distance, velocity, and displacement through the use of the motion data gathered with a cart simulation. This will be used to extrapolate and calculate graphs and equations for the positions and velocity.

EXPERIMENTAL DATA (3 points):

PART 1: Object moving away from the motion sensor.

Table 1

Table 2

Using equation 5, substitute in your slope and y-intercept to create an equation that describes the motion of the cart:

Run 1_y=0.5160x -3.376

Run 2: y=-0.00260x+0.5383

Table 3. (Use Logger Pro for statistical data to calculate avg. velocity and uncertainty)

PART 2: Object moving toward the motion sensor.

Table 4

Using equation 5, substitute in your slope and y-intercept to create an equation that describes the motion of the cart:

Run 3: Position = -.8242t + 3.108

Run 4: Position = -1.235t + 7.175

Table 5. (Use Logger Pro for statistical data to calculate avg. velocity and uncertainty)

PART 3. Matching position vs. time graphs

Run 5: Include in the Discussion the methods used to reproduce the graph.

Insert the best match for the position vs time graph.

DATA ANALYSIS (10 points): the section includes equations, calculations and error analysis if required. Be sure all equations are present!

PART 1 & 2:

Show the equation with the plugged in numbers you used in run 1 to calculate the cart’s:

·Distance: xFinal – xInitial = Distance traveled

1.70 m – .85 m = .85 m

(

·Displacement: xFinal – xInitial = Displacement

1.70 m – .85 m = .85 m

·Average speed: Displacement / Time = Average speed

.85 m / (9.843 – 8.187) s = .513 m/s

·Average velocity: Displacement / Time = Average velocity

.85 m / (9.843 – 8.187) s = .513 m/s

For run 2 and 4, calculate the percent difference between the average velocity from the slope of the x(t) graph and mean velocity from v(t) graph.

Let x=slope of x(t) & v=mean velocity of v(t)

% Difference = 100 * |x+v|/((x+v)/2)

Run #2: 100 * |.516+.5162|/((.516+.5162)/2) = 0.0387%

Run #4: 100 * |-1.235+-1.275|/((-1.235+-1.275)/2) = 3.187%

PART 4:

Run 6: Show calculation (use eqn. 5) used to make a prediction for the final position of the cart after 2 s from the moment it crosses the 80 cm mark on the track.

x(t) = x0 + v(t-t0)

x(t)= (80cm) + (-20m/s)(2s-0s)

x(t)=80cm +(-40cm)

x(t)=40cm

RESULTS (3 points):

Table 6. PARTS 1 & 2 (Report results below with 4 significant figures)

Table 7. PARTS 1 & 2 Equations of motion

Table 8. PART 4 final position of cart after 2 s

DISCUSSION AND CONCLUSION (10 points):