How Do You Calculate Velocity and Acceleration from a Cubic Position Function?

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To calculate velocity and acceleration from the cubic position function x = bt^3, where b = 1.5 m/s^3, the derivatives yield v = 3bt^2 and a = 6bt. At t = 2.5 seconds, the instantaneous velocity is 28.125 m/s and the instantaneous acceleration is 22.5 m/s^2. The average velocity over the first 2.5 seconds is 11.25 m/s, while the average acceleration is 9 m/s^2. It is clarified that instantaneous values differ from average values, with no averaging applied to instantaneous calculations.
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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?
 
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Your symbolic expressions for velocity and acceleration are fine, but where are your numerical answers ?
 
arent those the answer? do you mean that i must plug in 1.5 for b? and 2.5 for the other t?
 
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.
 
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?
 
ProBasket said:
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?

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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