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Homework Help: "speed rate of change" vs. "velocity rate of change"

  1. Dec 14, 2015 #1


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    1. The problem statement, all variables and given/known data
    At a moment in time a particle has a velocity of v = 3 m/s î + 4 m/s ĵ, and an acceleration of a = 6 m/s2 î + 3 m/s2 ĵ. Find the rate of change of the speed of the particle, that is, find d|v|/dt.
    (better formatted reference: click)​

    The answer is: 6 m/s2. How on Earth????

    2. Relevant equations
    speed = magnitude of velocity
    a = dv/dt​

    3. The attempt at a solution
    The magnitude of the initial velocity (the speed) is 5 m/s. 3^2 + 4^2 = 5^2.
    Acceleration is constant, so I tried using the equations of uniform acceleration kinematics, e.g. vx=∫(ax)dt, etc.
    Can't, for the life of me, see how the rate of change of speed = 6m/s2.
    Thanks for any hints.​
    Last edited: Dec 15, 2015
  2. jcsd
  3. Dec 15, 2015 #2


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    Homework Helper

    Write the velocity as function of time, determine the speed as function of time, then differentiate. That is the rate of change of speed
    at time t. You need the rate of change at t=0.
  4. Dec 15, 2015 #3


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    Hmm... trying to follow your advice.
    click here
    The stated answer is 6 m/s2, which I'm not finding any way to get. Maybe 6 m/s2 is incorrect?
  5. Dec 15, 2015 #4


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    The right hand side of the first line of step 6 is wrong. How do you differentiate ##\sqrt{x^2+y^2}##?
    An easier way is to start with ##|v|^2=\vec v.\vec v## and differentiate both sides.
  6. Dec 15, 2015 #5


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    Ah! Got it!! Thank you so much.
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