help please! instantaneous velocity and avarage velocity?

Question

The position of a small object is given by X=34+10t+2t^3, where t is in seconds and x in meters. X is a function of t when graphed t=0 to t=3.0.

what is the average velocity of the object between 0 and 3.0s? at what time between 0 and 3.0s is the instantaneous velocity zero?

Answer 1

1. instantaneous velocity - velocity as a function of time at a particular time for a particular displacement function.

The slope at a particular time along the displacement function is your instantaneous velocity

X=34+10t+2t^3 then
v(t)=X'=10+6t^2 is instantaneous velocity.

now compute v(0)=10 and v(3.0)=64


2. average velocity - displacement over time


Vavg=(X2-X1)/(t2-t1)
X1=34
X2=118
Vavg=(118-34)/(t3-0)=28


v(t)=X'=10+6t^2 is 0 at ? Time would have to be an imaginary number LOL

Answer 2

OK. So you take the derivative of x with respect to t to get the instantaneous velocity dx/dt. Then you have to look at that function for 0's (roots) which is where the volocity will be 0. Just set it equal to zero and solve it (it'll be a quadratic in x).
To get the average velocity, look at the position at t=0 and at t=3. Then (f(3)-f(0))/3 will be the average velocity (distance travelled divided by travel time).

HTH

Doug

Answer 3

X = 2t³ +10t+34
V =dX/dt =6t²+10

[dX/dt]@ t=0 is V0=10m/s
[dX/dt]@ t=3is V=54+10 =64

avg Velocity = (64-10)/3 = 54/3 = 18

inst. velocity = 0
ie; dX/dt = 0
6t²+10=0
3t²+5 = 0
t=±√(-5/3) =±i√(5/3) where i =√(-1)

ie; inst velocity is zero when t = i√(5/3)