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.