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Homework Help: Simple physics question

  1. Jan 30, 2005 #1
    The position of an object is given by x= bt^3, where x is in meters, t is in seconds, and where the constant b is 1.5 m/s^3. Determine (a) the instantaneous velocity and (b) the instantaneous acceleration at the end of 2.5s. Find (c) the average velocity and (d) the average acceleration during the first 2.5s.

    taking the derivative of x will give me the velocity and taking the derivative of velocity will get me the acceleration, so...

    x= bt^3
    v= 3bt^2
    a= 6bt

    here are my answers (please check them):

    a) bt^3/2.5
    b) 3bt^2/2.5
    c) bt^3/2.5
    d) 3bt^2/2.5

    can someone check these answers and tell me if i'm doing them correctly?
  2. jcsd
  3. Jan 30, 2005 #2


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    Homework Helper

    Your symbolic expressions for velocity and acceleration are fine, but where are your numerical answers ?
  4. Jan 30, 2005 #3
    arent those the answer? do you mean that i must plug in 1.5 for b? and 2.5 for the other t?
  5. Jan 30, 2005 #4


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    Yes, you should plug in the values.a) and b) are fairly direct. c) and d) are also simple, but you need to take a different tack. I want to see your answers before helping out.
  6. Jan 30, 2005 #5
    c) bt^3/2.5
    (1.5)(2.5)^3/2.5 = 9.375 m/s^3
    d) 3bt^2/2.5
    (3)(1.5)(2.5)^2/2.5=50.625 m/s^3

    would a.) and c.) have the same answer? and would b.) and d.) have the same answer?
  7. Jan 30, 2005 #6


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    What you're doing for c) and d) is correct. a) and b) would NOT in general have the same answer as c) and d) respectively.

    Average velocity taken over an interval of time is very simply (final displacement - initial displacement)/total time. Ave. acceleration is similarly (final velocity - initial velocity)/total time.

    Instantaneous velocity referst to the velocity of the particle at that instant of time. No averaging should be done. Ditto for the inst. acceleration.

    What are the numerical answers for the inst. velocity and accel ?
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