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

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Homework Help Overview

The problem involves calculating velocity and acceleration from a cubic position function, specifically given by x = bt³, where b is a constant. The discussion focuses on determining instantaneous and average values at a specific time of 2.5 seconds.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss taking derivatives of the position function to find velocity and acceleration. There is a query about whether to substitute numerical values for the constants and time in the derived expressions. Some participants seek clarification on the relationship between instantaneous and average values.

Discussion Status

Participants have provided symbolic expressions for velocity and acceleration and are in the process of substituting numerical values. There is an ongoing exploration of the differences between instantaneous and average quantities, with some guidance offered regarding the calculation methods for average velocity and acceleration.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance provided. There is a focus on ensuring understanding of the definitions of instantaneous versus average values.

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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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