Constant Velocity Lab: Part 1 ANSWER ALL QUESTIONS (FULL SENTENCES) Questions: 1

Question

Constant Velocity Lab:

Part 1

ANSWER ALL QUESTIONS (FULL SENTENCES)

Questions:

1. How do successive displacement of the cart compare? Why?

2. What is the slope of the position vs. time graph? What is the significance of this value?

3. How would the position vs. time graph be different if the cart had gone faster or slower?

4. What is the slope of the velocity vs. time graph? What is the significance of this value?

5. Was the velocity of the cart more-or-less constant during its motion? How do you know?

6. How would the velocity vs. time graph be different if the cart had gone faster or slower?

7. What was the acceleration of the cart during its motion? How do you know?

Part 2

State a Hypothesis

Part 3

Conclusion: (3 Paragraphs)

1. Quantitative: Address the purpose and results

2. Sources of error

3. Suggestions for improvement

To study the motion of a motorized cart. Motion including its position, displacement, velocity and acceleration. Also to practice constructing position vs. time graphs for a motion. Hypothesis: Procedure: Use a constant motion machine and a ticker timer to collect data. Data Table Time (s) (displacement) a (velocity) Cm v (cm/s) Ad (cm) 0.000 3.65 3.65 0.100 36.5 0.200 3.83 7.48 38.3 3.72 37.2 11.20 0.300 378 0.400 14.98 3.78 378 18.76 3.78 0.500 389 0.600 3.89 22.65 39.0 0.700 26.55 3.90 3.82 0.800 30.37 38.2 0.900 34.38 4.01 40.1 3.07 30.7 37.45 1.00

Explanation / Answer

1) Successive displacement are almost equal and comparable. Fo constant velocity lab, the successive displacments will be small.

2) The slope of position vs time graph gives velocity. The value of slope tells us how fast or slow an object is moving.  If the velocity is constant, then the slope is constant (i.e., a straight line). If the velocity is changing, then the slope is changing (i.e., a curved line). If the velocity is positive, then the slope is positive.

3) If there is an change in velocity, the slope of position vs time graph will be a curve ( going upward for acceleration and going downward for deceleration)

4) The slope of velocity vs time graph gives acceleration. it tells us at what rate the velocity is changing in a given interval of time.

5) No, veloctiy was having a difference of 25 % as calculated by you

6) same explanation as position vs time graph

7) a = v2 - v1 / t2 - t1

a = 38.3 - 36.25 / 0.2 - 0.1

a = 18 m/s2

Note this is with respect to first two readings in your table

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