Stupid Ticker Tape 2

Topics: Acceleration, Velocity, Kinematics Pages: 3 (512 words) Published: April 13, 2015
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...
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