Answer to Question #162790 in Mechanics | Relativity for ashelu

Question

Purpose:

The purpose of this activity is to determine the acceleration of a falling object, through manual measuring and graphing techniques, using the object’s position at various times.

Table 1 – Object’s Position

Time (s) position (m(down))

0.0 0.50

2.0 0.70

4.0 1.00

6.0 1.25

8.0 1.60

10.0 2.00

Instructions:

1. Plot the points above into a position-time graph.

2. Draw a curved line-of-best-fit through the plotted points.

3. Draw a minimum of 3 tangents and determine the slope for each. Show all calculations and include units.

4. Create a new table of values using time and instantaneous velocity as the variables.

5. Plot these points into a velocity-time graph and draw a best fit line.

6. Determine the slope of the line and show all work.

7. Draw a third graph showing the acceleration of the vehicle for 10 seconds. Shade the area under the line.

Discussion:

1. What was the graphically determined acceleration?

2. Calculate the area under the line of the V-T and A-T graph and indicate what it represents.

3. What are the limitations of using this technique for determining acceleration?

(need to do the instructions part and discussion part)

Answer 1

The plotted graph is given below as per the given table,

Slop of the graph

$v_1 = \frac{0.7-0.5}{2}=0.1m/s$

$v_2=\frac{1-0.7}{4-2}=0.15m/s$

$v_3=\frac{1.25-1}{2}=0.125m/s$

$v_4=\frac{1.6-1.25}{2}=0.175m/s$

$a_1=\frac{v_2-v_1}{2}=\frac{0.05}{2}=0.025m/s^2$

$a_2=\frac{v_3-v_2}{2}=\frac{0.125-0.15}{2}=\frac{-0.025}{2}=-0.0125m/s^2$

$a_3=\frac{v_4-v_2}{2}=\frac{0.175-0.125}{2}=\frac{0.05}{2}=0.0125 m/s^2$