Answer: v = 3t^2, -e^-t ;
Speed = √ 9t^4 + (-e^-t)^2 ;
a(t) = 6t, e^-t
Explanation:
Given the position function at time 't' for t greater than or equal to 0:
r(t) = (9,t^3, e^-t)
The Velocity of 'v' is the first derivative of r with respect to 't' :
d/dt(9) = 0
d/dt(t^3) = 3t^3-1 = 3t^2
y = (e^-t) =
Using chain rule :
Let u = -t
y = e^u ; dy/du = e^u
du/dt = -1 ;
Therefore dy/dt = dy/du * du/dt
dy/dt = -1 * e^u ; -e^u ; u = -t
dy/dt = -e^-t
Therefore,
v = 3t^2, -e^-t
The magnitude of velocity 'v ' = speed
Speed = √v1^2 + v2^2
Speed = √(3t^2)^2 + (-e^-t)^2
Speed = √ 9t^4 + (-e^-t)^2
Acceleration(a) = derivative of velocity
d/dt(3t^2) = 6t
d/dt (-e^-t) :
Using chain rule :
dy/dt = -e^-t
Let u = -t ; du/dt = -1
-e^u ; dy/du = -e^u
Therefore, dy/dt = -1 * -e^u
u = -t
dy/dt = e^-t
Therefore, acceleration:
a(t) = 6t, e^-t