Stupid Ticker Tape 2

Science 10. P 393 Investigation 13A
Time (sec)
Displacement (cm down)
Velocity ( down)
0
0

0
0.1
0.6
= Df – Di = 0.6 – 0 = 0.6 =
0.2
1
= Df – Di = 1 – 0.6 = 0.4 =
0.3
3.3
= Df – Di = 3.3 – 1 = 2.3 =
0.4
5.9
= Df – Di = 5.9 – 3.3 = 2.6 =
0.5
7.4
= Df – Di = 7.4 – 5.9 = 1.5 =
0.6
8.7
= Df – Di = 8.7 – 7.4 = 1.3 =
0.7
10.1
= Df – Di = 10.1 – 8.7 = 1.4 =
0.8
11.4
= Df – Di = 11.4 – 10.1 = 1.3 =
0.9
12.9
= Df – Di = 12.9 – 11.4 = 1.5 =
1.0
11.8
= Df – Di = 11.8 – 12.9 = -1.1 =
1.1
10.5
= Df – Di = 10.5 – 11.8 = -1.3 =
Length of ramp: 80 cm
Total distance: 87.5 cm
Average velocity: vav = (as it was a straight line the distance is equal to displacement)= down =

a) Slopes (Of Lines a, b, & c):
a. a = Δv = vf – vi = – 0 = Δt = tf – ti = 0.1s – 0s = 0.1s a = = (Acceleration)
b. a = Δv = vf – vi = – = Δt = tf – ti = 0.525s – 0.4s = 0.125s a = = (Deceleration)
c. a = Δv = vf – vi = – = Δt = tf – ti = 1.1s – 0.55s = 0.55s a = = (No acceleration)
b) Questions a => g
a. Our line of best fit (in each section, in this case section a) is a straight line. This indicates that there was constant positive acceleration
b. a = Δv = vf – vi = – 0 = Δt = tf – ti = 1.1s – 0s = 1.1s a = = (Acceleration)
c. (Equal to the slope as the slope in a velocity-time graph, the slope is equal to acceleration)
d. Constant acceleration is shown by a straight line
e. Our line was straight. However, if it was not straight, that may have been due to friction
f. Yes; whenever the acceleration was constant, it would be a straight line. When there is no acceleration, it is still technically constant, so therefore there would also be a straight line.
g. There would have been greater acceleration (steeper slope) if the ramp had been made When the angle is greater/steeper this causes gravity to have a greater