The position of an object as a function of time is given by x = At2 – Bt + C, where A = 7.3 m/s2, B = 5.8 m/s, and C = 4.5 m. Find the instantaneous velocity and acceleration as functions of time. (Use the following as necessary: t.)

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v=dx/dt=2At-B

a=dv/dt=2A

To find the instantaneous velocity and acceleration as functions of time, we need to take the derivative of the position function with respect to time.

Step 1: Find the derivative of x = At^2 - Bt + C with respect to time t.

The derivative of At^2 with respect to t is 2At, since the derivative of t^2 is 2t.

The derivative of -Bt with respect to t is -B, since the derivative of t is 1.

The derivative of C with respect to t is 0, since C is a constant.

So, the derivative of the position function x = At^2 - Bt + C is dx/dt = 2At - B.

Step 2: The derivative dx/dt represents the instantaneous velocity, so we can express the instantaneous velocity as a function of time.

Therefore, the instantaneous velocity v(t) = 2At - B, where A = 7.3 m/s^2 and B = 5.8 m/s.

Step 3: To find the acceleration function, we take the derivative of velocity v(t) with respect to time t.

The derivative of 2At with respect to t is 2A, since the derivative of t is 1.

The derivative of -B with respect to t is 0, since B is a constant.

So, the derivative of the velocity function v(t) = 2At - B is dv(t)/dt = 2A.

Step 4: The derivative dv(t)/dt represents the instantaneous acceleration, so we can express the instantaneous acceleration as a function of time.

Therefore, the instantaneous acceleration a(t) = 2A, where A = 7.3 m/s^2.

In summary:
- The instantaneous velocity v(t) = 2At - B, where A = 7.3 m/s^2 and B = 5.8 m/s.
- The instantaneous acceleration a(t) = 2A, where A = 7.3 m/s^2.

To find the instantaneous velocity and acceleration as functions of time, we need to differentiate the position function with respect to time.

Given: x = At^2 - Bt + C

Instantaneous velocity, v(t), is given by the derivative of position with respect to time:

v(t) = dx/dt

Differentiating the position function x = At^2 - Bt + C with respect to t, we get:

dx/dt = d/dt (At^2 - Bt + C)
= 2At - B

Therefore, the instantaneous velocity as a function of time is:

v(t) = 2At - B

Now, to find the acceleration, a(t), we need to differentiate the velocity function with respect to time:

a(t) = dv/dt

Differentiating the velocity function v(t) = 2At - B with respect to t, we get:

dv/dt = d/dt (2At - B)
= 2A

Therefore, the instantaneous acceleration as a function of time is:

a(t) = 2A

Substituting the given values A = 7.3 m/s^2 into the equations:

v(t) = 2(7.3t) - 5.8
= 14.6t - 5.8 m/s

a(t) = 2(7.3)
= 14.6 m/s^2

So, the instantaneous velocity as a function of time is v(t) = 14.6t - 5.8 m/s,
and the instantaneous acceleration as a function of time is a(t) = 14.6 m/s^2.