Help with average velocity during a time interval

[dealdoe22]

Homework Statement

The position of a particle in meters moving along the x-axis is given by: x=21+22t-6t^2
Calculate the average velocity of the object in m/s during the time interval t=1s to t=3s.

Homework Equations

x=21+22t-6t^2
t=3 ; t=1
v avg. = Xf - Xi/Tf - Ti = Δx/Δt

don't know if this applies but:
instantaneous velocity = dx/dt = -12t+22

The Attempt at a Solution

Hello all. I am a returning student with a Major in Civil Engineering. At the age of 35, it is difficult to remember sometimes what the simple answers are. I know this may seem elementary to some of you but I believe I am over thinking this question. Please help.

My question is which answer is correct for the question asked?
I attempted this 2 different ways and can't decide if one of them is correct or if either of them are correct.

substituting:
x=21+22(3)-6(3)^2
x=33m
x=21+22(1)-6(1)^2
x=37m

then 33-37/3-1 = -2 m/s for velocity avg.
speed = absolute value of velocity = 2 m/s

or

I took the derivative of the initial equation and then substituted to get:
v=-12t+22
v=-12(3)+22=-14
v=-12(1)+22=10

then -14-10/3-1=-12m/s = 12m/s

Any help would be great. Thank you.

 

[PhanthomJay]

Average velocity is total displacement divided by total time, so your first method is the correct way to proceed, v_ave= -2 m/s. Note, however, that you can't just take the absolute value of velocity to determine the average speed, this is incorrect, since ave speed is total distance over total time, and the total distance traveled is greater than the total displacement.

 

[dealdoe22]

Excellent. Thank you very much. I was over thinking this one. Just a quick question to confirm what the derivative is used for.

I believe it is used to find instantaneous velocity. So, if the question asked me to find the velocity at a certain point, t=3, then I would substitute 3 for t into the derivative and calculate?

Thank you for your help again. It is much appreciated.

 

[PhanthomJay]

Excellent. Thank you very much. I was over thinking this one. Just a quick question to confirm what the derivative is used for.

I believe it is used to find instantaneous velocity. So, if the question asked me to find the velocity at a certain point, t=3, then I would substitute 3 for t into the derivative and calculate?

Yes, that is correct. I also want to point out that for constant acceleration, the average velocity can also be found by your alternate method, using V_avg =( V_i + V_f)/2 (although you mistakenly added a minus sign), that is, V_avg = (-14 + 10)/2 = -2 m/s, which yields the same result. You know you have constant acceleration in this problem by differentiating the velocity equation, a = dv/dt = -12 m/s^2, constant acceleration. However, this alternate method does NOT apply for the general case when acceleration is not constant, so it is best to use the first approach which always works when you want to find average velocity, unless you are sure that the acceleration is constant, in which case you can use either method.