The motion of the particle is described by the equation x(t)= -t^2+Bt+C where B and C are constants (not necessarily positive). Derive equations for both the velocity v and the acceleration a of the particle as functions of time. Express v and a in terms of B, C and t, as appropriate

Given the position function \( x(t) = -t^2 + Bt + C \), we can derive the velocity and acceleration as follows:

### 1. Velocity \( v(t) \):
Velocity is the rate of change of position with respect to time, which is the first derivative of \( x(t) \) with respect to \( t \).

\[
v(t) = \frac{dx(t)}{dt}
\]

Taking the derivative of \( x(t) \):

\[
v(t) = \frac{d}{dt} \left( -t^2 + Bt + C \right)
\]

Using basic differentiation rules:

- \( \frac{d}{dt}(-t^2) = -2t \)
- \( \frac{d}{dt}(Bt) = B \)
- \( \frac{d}{dt}(C) = 0 \) (since \( C \) is constant)

Thus,

\[
v(t) = -2t + B
\]

### 2. Acceleration \( a(t) \):
Acceleration is the rate of change of velocity with respect to time, which is the first derivative of \( v(t) \) or the second derivative of \( x(t) \).

\[
a(t) = \frac{dv(t)}{dt} = \frac{d^2x(t)}{dt^2}
\]

Taking the derivative of \( v(t) = -2t + B \):

\[
a(t) = \frac{d}{dt} \left( -2t + B \right)
\]

Since \( \frac{d}{dt}(-2t) = -2 \) and \( \frac{d}{dt}(B) = 0 \), we have:

\[
a(t) = -2
\]

### Final expressions:
- Velocity: \( v(t) = -2t + B \)
- Acceleration: \( a(t) = -2 \)

These equations describe the motion of the particle, where the velocity depends on both time \( t \) and the constant \( B \), while the acceleration is constant at \( -2 \).

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Do you have any questions, or would you like further details on any step?

Here are 5 related questions to extend this topic:
1. How would the motion change if \( B \) were negative?
2. What would the particle's velocity be at \( t = 0 \)?
3. How do the values of \( B \) and \( C \) affect the initial conditions of the motion?
4. How would you interpret a scenario where \( B = 0 \)?
5. What is the significance of the constant acceleration in this motion?

**Tip**: The derivative of a function provides critical insights into its rate of change. In physics, velocity is the derivative of position, and acceleration is the derivative of velocity.

The motion of the particle is described by the equation x(t) = -t^2 + Bt + C where B and C are constants (not necessarily positive). Derive equations for both the velocity v and the acceleration a of the particle as functions of time. Express v and a in terms of B, C, and t, as appropriate.

Learn how to derive the velocity and acceleration equations for a particle moving according to x(t) = -t^2 + Bt + C. This problem covers key concepts in calculus and kinematics, suitable for students in Grades 11-12.

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