A cart is moving along a straight track. The cart's position x is initially zero, and its velocity v asna function of time t is described by the equation v(t) = bt^2 - ct, where b and c are positive constants. The cart is instantaneously at rest at time t = tR      a) Sketch a graph of the cart's acceleration a as a function of t for the time interval shown      b) Derive an expression for the cart's position xR that is negative and the farthest away from the initial position (x = 0)           c) Derive an expression for the velocity v0 that the cart has when it passes through the position x = 0 at a later time t > 0

Let's tackle each part of this problem step by step.

### Part (a): Sketching the Acceleration \( a(t) \) as a Function of Time \( t \)

The velocity of the cart is given by the equation:
\[
v(t) = bt^2 - ct
\]
where \( b \) and \( c \) are positive constants.

To find the acceleration \( a(t) \), we take the derivative of \( v(t) \) with respect to \( t \):
\[
a(t) = \frac{dv(t)}{dt} = \frac{d}{dt}(bt^2 - ct) = 2bt - c
\]

So, the acceleration as a function of time is:
\[
a(t) = 2bt - c
\]

This is a linear equation, meaning the graph of \( a(t) \) versus \( t \) is a straight line with slope \( 2b \) and y-intercept \( -c \).

- When \( t = 0 \), \( a(0) = -c \) (negative acceleration).
- As time increases, \( a(t) \) increases linearly.
- The acceleration becomes zero when \( 2bt = c \), i.e., at \( t = \frac{c}{2b} \).
- For \( t > \frac{c}{2b} \), the acceleration is positive.

### Part (b): Deriving the Expression for \( x_R \)

To find the position \( x_R \) where the cart is farthest from the initial position (and \( x_R \) is negative), we first need to find the time \( t_R \) when the cart is at rest, which happens when \( v(t) = 0 \).

Set \( v(t) = 0 \):
\[
0 = bt_R^2 - ct_R
\]
\[
t_R(bt_R - c) = 0
\]
This gives us two solutions:
1. \( t_R = 0 \) (initial condition, which we ignore since we want the non-zero time)
2. \( t_R = \frac{c}{b} \)

The position \( x(t) \) is found by integrating the velocity function:
\[
x(t) = \int v(t) \, dt = \int (bt^2 - ct) \, dt = \frac{b}{3}t^3 - \frac{c}{2}t^2 + C
\]
Since \( x(0) = 0 \), the constant \( C \) is zero:
\[
x(t) = \frac{b}{3}t^3 - \frac{c}{2}t^2
\]

At \( t = t_R = \frac{c}{b} \), we find \( x_R \):
\[
x_R = x\left(\frac{c}{b}\right) = \frac{b}{3}\left(\frac{c}{b}\right)^3 - \frac{c}{2}\left(\frac{c}{b}\right)^2
\]
\[
x_R = \frac{c^3}{3b^2} - \frac{c^3}{2b^2}
\]
\[
x_R = \frac{c^3}{b^2} \left(\frac{1}{3} - \frac{1}{2}\right) = -\frac{c^3}{6b^2}
\]

So, the position \( x_R \) where the cart is farthest from the initial position is:
\[
x_R = -\frac{c^3}{6b^2}
\]

### Part (c): Deriving the Expression for \( v_0 \) When the Cart Passes Through \( x = 0 \) at a Later Time \( t > 0 \)

The cart will pass through \( x = 0 \) again at some later time \( t_0 > 0 \). To find \( t_0 \), we set the position equation to zero:
\[
0 = \frac{b}{3}t_0^3 - \frac{c}{2}t_0^2
\]

This simplifies to:
\[
t_0^2\left(\frac{b}{3}t_0 - \frac{c}{2}\right) = 0
\]
Ignoring \( t_0 = 0 \), we have:
\[
t_0 = \frac{3c}{2b}
\]

Now, substitute this back into the velocity equation to find \( v_0 \) (the velocity when the cart passes through \( x = 0 \) again):
\[
v_0 = v\left(\frac{3c}{2b}\right) = b\left(\frac{3c}{2b}\right)^2 - c\left(\frac{3c}{2b}\right)
\]
\[
v_0 = \frac{9c^2}{4b} - \frac{3c^2}{2b}
\]
\[
v_0 = \frac{9c^2 - 6c^2}{4b} = \frac{3c^2}{4b}
\]

Thus, the velocity \( v_0 \) when the cart passes through \( x = 0 \) again is:
\[
v_0 = \frac{3c^2}{4b}
\]

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Do you want more details or have any questions?

Here are five related questions to explore further:
1. How would the graph of \( a(t) \) change if \( b \) were negative?
2. What would happen to the cart’s motion if \( c = 0 \)?
3. How would the maximum negative displacement \( x_R \) change if the mass of the cart were taken into account?
4. Can you find the time at which the cart first returns to \( x = 0 \)?
5. How would air resistance affect the velocity and position functions?

**Tip:** When dealing with motion problems, always consider starting by analyzing the velocity and acceleration functions, as they give direct insights into the nature of the motion.

Explore the dynamics of a cart moving along a straight track with a detailed analysis of acceleration, position, and velocity equations. Includes graphical representations and derivations for key points in the motion.

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