Physics Help: Interpreting Position-versus-Time graphs?

“Find the object’s speeds v1, v2, and v3 at times t1 = 2.0s, t2 = 4.0s, and t3 = 13s. Express your answers in meters per second to two significant figures, separated by commas.”

There is a graph given, but I clearly can’t copy and paste it here and I can’t just give the link to the graph because MasteringPhysics.com (the site used for my homework) requires a password. But I’ve already triple checked and the positions of the object are 4m at t1, 7m at t2, and 9m at t3.

To find the velocity, don’t I just take the change in position over the change in time? If so, I get the answers 2.0, 1.75, and .69, but Mastering Physics tells me those answers are wrong….

Any suggestions? I’m desperate!

Thank you in advance!!

Anonymous: I completely understand your answer, but Mastering Physics apparently doesn’t…I tried “2.0,1.5,.22” as my answer and it still says that it’s incorrect 🙁 Any other suggestions? I have 4 attempts at the answer left, but I do lose a percentage each time I guess incorrectly. Thank you for your help so far!

5 Answers

  • You need to use the formula Velocity = Distance / Time.

    Since it started from 0 you can say your change in position for t1 is 4 meters. So it’s 4 meters / 2 seconds for t1. You get 2 m/s.

    For t2 you go from t1 to t2 a total of 3 meters. So it’s 3 meters / 2 seconds. So it’s 1.5 m/s.

    For t3 you go from t2 to t3 a total of 2 meters. It takes 9 seconds. So it’s 2 meters / 9 seconds which is .22 m/s (approximately)

    See if that works. The question should say if it wants between certain time intervals or the average velocity.

  • The velocity is the slope of a displacement vs time graph.

    At t = 0 s, x = 0 m, at t = 2 s, x = 4 m

    If the graph, for the 1st period of time, is a straight line, slope = ∆x/∆t = 4/2 = 2

    Average velocity during this time is 2 m/s.

    During the next 2 seconds, the object moves from 4 m to 7 m.

    ∆x = 3 m, ∆t = 2 s,

    If the graph, for the 2nd period of time, is a straight line, slope = 3/2 = 1.5 m/s

    Average velocity = 1.5 m/s

    During the next 9 seconds, the object moves from 7 m to 9 m

    If the graph for the 3rd period of time is a straight line, slope = 2/9 = 0.222 m/s

    Average velocity = 0.222 m/s

  • Find Slope for each one:

    t1= 7/3 = 2.3m/s

    t2= 0 =0m/s

    t3= 3/5 =0.60m/s

  • Times T1

  • On a place vs time graph the cost of the object is given via the magnitude of the lines slope. speed incorporate the magnitude of the slope and no count if it is helpful or unfavourable.

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