ReviewEssays.com - Term Papers, Book Reports, Research Papers and College Essays
Search

Speed and Acceleration of Cars on Lambert Road

Essay by   •  December 2, 2010  •  Research Paper  •  2,314 Words (10 Pages)  •  1,605 Views

Essay Preview: Speed and Acceleration of Cars on Lambert Road

Report this essay
Page 1 of 10

Speed and Acceleration of Cars on Lambert Road

Aim: To measure the speeds of cars traveling on and the acceleration of cars as they stop a red light on Lambert Road.

Hypothesis: Most cars traveling on Lambert road will travel at a legal speed of 60km/hr or under. Cars will start accelerating at exactly the same distance before the lights and stop at exactly the white line in front of the lights.

Hazards: - being hit by a car

- falling onto the road

Materials: - A rope with 5m intervals

- A stopwatch

Method: Part 1: Speed

1. A distance of 20m was measured using the rope. One group member was placed at the start and one at the finish.

2. As a car passed the start, the stopwatch was started.

3. As the car reached the finish, the group member at the finish made a shouting signal and the stopwatch was stopped.

4. The times were recorded and the speed calculated using the formula v=s/t.

5. Steps 1-4 were repeated with nine more cars.

Part 2: Acceleration

1. A distance of 50m was measured using the rope. One group member was placed at the start and one at the finish.

2. As a car passed the start, the stopwatch was started.

3. As the car reached the finish, the group member at the finish made a shouting signal and the stopwatch was stopped.

4. The times were recorded and the acceleration calculated using the formula a = s ч (vt - Ð... tІ).

5. Steps 1-4 were repeated with four more cars.

Results: First set

Speed

Car Distance - s (m) Time - t (s) Speed - v (m/s) v= s/t Speed - v

(km/hr)

1 20.0 ± 0.5 2.22 9.1 32.7

2 20.0 ± 0.5 1.41 14.2 51.1

3 20.0 ± 0.5 1.28 15.6 56.3

4 20.0 ± 0.5 1.44 13.9 50.0

5 20.0 ± 0.5 1.47 13.6 50.0

6 20.0 ± 0.5 1.28 15.6 56.3

7 20.0 ± 0.5 1.28 15.6 56.3

8 20.0 ± 0.5 1.85 10.8 38.9

9 20.0 ± 0.5 1.44 13.9 50.0

10 20.0 ± 0.5 1.31 15.3 55.0

Acceleration

Car Distance - s (m) Final speed - v (m/s) Time - t (s) Acceleration - a (m/sІ)

a = s ч (vt - Ð... tІ)

1 50.0 ± 0.5 0 12.65 -0.6

2 50.0 ± 0.5 0 8.84 -1.3

3 50.0 ± 0.5 0 8.41 -1.4

4 50.0 ± 0.5 0 9.87 -1.0

5 50.0 ± 0.5 0 10.69 -0.9

Results: Second set

Speed

Car Distance - s (m) Time - t (s) Speed - v (m/s) v= s/t Speed - v

(km/hr)

1 20.0 ± 0.5 1.30 15.4 55.4

2 20.0 ± 0.5 1.45 13.8 49.6

3 20.0 ± 0.5 1.37 14.6 52.6

4 20.0 ± 0.5 1.19 16.8 60.5

5 20.0 ± 0.5 1.53 13.1 47.1

6 20.0 ± 0.5 1.44 13.9 50.0

7 20.0 ± 0.5 1.42 14.1 50.7

8 20.0 ± 0.5 1.28 15.6 56.3

9 20.0 ± 0.5 1.30 15.4 55.4

10 20.0 ± 0.5 1.57 12.7 45.9

Acceleration

Car Distance - s (m) Final speed - v (m/s) Time - t (s) Acceleration - a (m/sІ)

a = s ч (vt - Ð... tІ)

1 50.0 ± 0.5 0 5.28 -3.6

2 50.0 ± 0.5 0 6.68 -2.2

3 50.0 ± 0.5 0 6.55 -2.3

4 50.0 ± 0.5 0 5.58 -3.2

5 50.0 ± 0.5 0 9.07 -1.2

Calculations: Working out speed (v):

v = s/t

=20/2.2

= 9.09

= 9.1 m/s

Working out acceleration (a):

s = vt - Ð... atІ

a = s ч (vt - Ð... tІ)

= 50 ч (0 x 8.84 - Ð... 8.84І)

= 50 ч -39.07

= -1.28

= -1.3 m/sІ

Converting m/s to km/hr:

m/s x 3.6

= 16.8 m/s x 3.6 = 60.5km/hr

Working out time (t):

s = 50.0

v = 0

u = 9.0 m/s

t = s ч (u + v)

2

= 50.0 ч (9.0 + 0)

2

= 50.0 ч (9.0)

2

= 50.0 ч 4.5

= 11.11

= 11s

s = 50.0

v = 0

u = 14.0

t

...

...

Download as:   txt (13.4 Kb)   pdf (162.5 Kb)   docx (15.1 Kb)  
Continue for 9 more pages »
Only available on ReviewEssays.com