Question

5. Lab Activity:

The animation video shows a car moving to the right with constant acceleration. I will run the car for different distances and pause. You measure the velocity and time each time, and record them in the table below.

Trial	Velocity (m/s)	Time, t (s)
1	5.02 m/s	3.350 s
2	6.52 m/s	4.350 s
3	7.87 m/s	5.249 s
4	8.85 m/s	5.960 s
5	10.20 m/s	6.799 s
6	12.07 m/s	8.049 s
7	12.82 m/s	8.549 s

a. Draw a graph with velocity (v) values on the y-axis and time (t) values on the x-axis.

b. Choose any two points on the graph and measure its slope.

v = m/s

t = s

Slope = m/s^2

c. What is the unit of the slope?

d. What is the relation between the acceleration of the car and the slope of the graph?

Answer 1

A) I have used Microsoft Excel to plot this simple graph.

B) One can directly get slope of the graph from MS Excel but we will compute that manually here.

Use following formula: Slope = dy/dx = (y2-y1)/(x2-x1)

use any two points for this calculation. Lets use 2nd and 6th point. so

slope = (12.07-4.35)/(8.049-6.52) = 5.55/3.699 = 1.5004 m/s²

C) As one can see from part B, unit of the slope is m/s²

D) Slope of the graph is the acceleration of the car. since, acceleration = ratio of chamge in velocity with respect to change in time i.e. dV/dt