The position versus time graph at right represents the motion of an object movin

Question

The position versus time graph at right represents the motion of an object moving in a straight line. a. Describes the motion. During which periods of time, if any, is the velocity constant? Explain how you can tell. b. Find the object's instantaneous velocity at each of the following times. Show your work. i. t = 0.5 s ii. t = 2.0 s iii. t=4.0 s How does the method you used to answer parts i-iii rely on your answer to ? c. For each of the following intervals, find the average velocity of the object. i. between A and C ii. between A and D iii. between B and D On the graph above, sketch and label the lines that would represent an object moving with constant velocity between each of the pairs of points in parts i-iii. For each line that you drew, how does the slope compare to the average velocity that you computed above? d. In which of the cases from part c, if any, is the average velocity over an interval equal to the average of the constant velocities occurring in that interval? [For example, is upsilon_AC (the average velocity from A to C) equal to 1/2(upsilon_AB + upsilon_BC)?|]

Explanation / Answer

we know that velocity = dx/dt = slope of any line segment in this graph so

average velocity =total dispalcemnt / total time

because the particle is moving in one direction so its displacement and distance will be same in any time interval

VAC = ( XC -XA) / t = (8-2) / 3 = 6/3 = 2 m/s

VAD = 10-2 / 5 = 8/5 = 1.6 m/s

VBD = (10-2 ) / 4 = 2 m/s

constant velocity means constant slope so

between

i = AB

Velocity = 0 contant

ii

BC B( 1 , 2) C ( 3 , 8 )

velocity = ( 2-8) / ( 1-3) = 6/2 = 3 m/s

iii C( 3 ,8 ) D ( 5 ,10)

V = (8-10) / ( 3-5) = -2 / -2 = 1 m/s

last part

in part Vac and Vbd both average speed and average velocity are same

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