Suppose the position of an object at time t is

(6 + 5t, 2t2, 7t − 5t3).
(a) What is the instantaneous velocity at time t?
(b) What is the acceleration at time t?
(c) What is the instantaneous velocity at time
t = 0?
(d) What is the acceleration at time
t = 0?

I know that v=dr/dt= d/dt(x,y,z). I do not understand how to find the instantaneous velocity without a line or time.
I also know that the instantaneous acceleration a=dv/dt.

To find the instantaneous velocity and acceleration at a given time, we need to take the derivatives of the position vector with respect to time.

(a) The instantaneous velocity at time t is given by v = dr/dt = (dx/dt, dy/dt, dz/dt).

To find dx/dt, take the derivative of the x-component of the position vector with respect to time:
dx/dt = d/dt(6 + 5t) = 5

To find dy/dt, take the derivative of the y-component of the position vector with respect to time:
dy/dt = d/dt(2t^2) = 4t

To find dz/dt, take the derivative of the z-component of the position vector with respect to time:
dz/dt = d/dt(7t - 5t^3) = 7 - 15t^2

So, the instantaneous velocity at time t is (5, 4t, 7 - 15t^2).

(b) The acceleration at time t is given by a = dv/dt.

To find dv/dt, take the derivative of the velocity vector with respect to time:
dv/dt = d/dt(5, 4t, 7 - 15t^2) = (0, 4, -30t)

So, the acceleration at time t is (0, 4, -30t).

(c) To find the instantaneous velocity at time t = 0, substitute t = 0 into the expression for velocity:
v(t=0) = (5, 4(0), 7 - 15(0)^2) = (5, 0, 7)

So, the instantaneous velocity at time t = 0 is (5, 0, 7).

(d) To find the acceleration at time t = 0, substitute t = 0 into the expression for acceleration:
a(t=0) = (0, 4, -30(0)) = (0, 4, 0)

So, the acceleration at time t = 0 is (0, 4, 0).