PHY-150 M2 Kinematics Lab Report.docx

Kinematics Bharat Sarma SNHU Margaret DeAntoni (m.deantoni@snhu.edu) Jan 21 st , 2024

Activity 1: Graph and interpret motion data of a moving object Activity 1. Table 1 Time (x axis) (seconds) Position (y axis) (meters) 0 0 5 20 10 40 15 50 20 55 30 60 35 70 40 70 45 70 50 55 Insert your graph here for Distance vs Time of a Moving Object. 0 10 20 30 40 50 60 0 10 20 30 40 50 60 70 80 Distance Vs Time Times in Seconds DIstance in Meters Questions for Activity 1 Question 1: What is the average speed of the train during the time interval from 0 s to 10 s? The Average speed = total distance /total time which is 40/10m = 4m/s Question 2:

Using the equation: v = s 2 − s 1 t 2 − t 1 , calculate the average speed of the train as it moves from position x = 50m to x = 60m. The v = s 2 − s 1 t 2 − t 1 , = (60-50)/(30-15) = 10/15 = 0.67m/s. Question 3: What does the slope of the line during each time interval represent? The time graph position in the slop represents the velocity during each time interval. Also, slop of the line represents acceleration and deceleration. Question 4: From time t = 35 s until t = 45 s, the train is located at the same position. What is slope of the line while the train is stationary? The slop = 0 m/s, when train is in rest however slop is 0. Question 5: Calculate the average speed of the train as it moves from position x = 70m to x = 55m. What does the sign of the average velocity during this time interval represent? v = x 2 − x 1 t 2 − t 1 = (55-70)/(50-35) = -15/15 = -1m/s. Which is represent the average displacement. Question 6: What is the displacement of the train from time t = 0s until t = 50s? Its shows the body changed its direction into negative sign. Question 7: What is the total distance traveled by the train from time t = 0s until t = 50s? The total distance travel by train is = 70m + 15m which is 85m. Question 8. What is the slope of the line during the time interval t = 45 to t = 50? Slop is = (55-70/(50-45) which is -3m/s. Question 9: What does the sign of the slope in question 8 represent in terms of the motion of the train? The question number 8 represent the change in direction of the objects with negative sign also negative velocity.

Question 10: What is the average velocity of the train during the interval t= 0s to t = 50s? The Average velocity v = s 2 − s 1 t 2 − t 1 , = (55 – 0)/(50 – 0) = 55/50 = 1.1m/s. Question 11: Does the train’s average velocity during the interval t= 0s to t = 50 s provide a complete picture of the train’s motion during this time? I say No, because here the average velocity is positive however in some time interval, body is in rest. The body also changing direction. The train travel 85m of total distance and with being stopped at 70m. Activity 2. Calculate the velocity of a moving object. Activity 2. Table 1 Time (s) Displacement (m)* 0.00 0.00 0.64 0.25 1.15 0.50 1.67 0.75 2.27 1.00 2.81 1.25 3.36 1.50 3.96 1.75 4.56 2.00 *Note that 0.25 m = 25 cm Insert a graph of Table 1 here. Include a chart title, axes titles and units. The table of graph is activity 2 table 1 which is Displacement(m) over Time(s).

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