To derive an equation for the car's velocity $v(t)$ as a function of time $t$, we start by recognizing that acceleration is the time derivative of velocity. The relationship between acceleration and velocity is given by:

$a(t) = \frac{dv}{dt}$

Given the acceleration function $a(t) = P - Qt$, we can integrate this with respect to $t$ to find the velocity as a function of time.

Step 1: Write down the differential equation

$\frac{dv}{dt} = P - Qt$

Step 2: Integrate both sides with respect to time

To find $v(t)$, we integrate the right-hand side with respect to $t$:

$v(t) = \int (P - Qt) \, dt$

Perform the integration:

$v(t) = P t - \frac{Q t^2}{2} + C$

where $C$ is the constant of integration.

Step 3: Apply the initial condition

We are told that the car starts from rest at $t = 0$, which implies that the initial velocity is zero:

$v(0) = 0$

Substitute $t = 0$ into the equation for $v(t)$:

$0 = P(0) - \frac{Q(0)^2}{2} + C \implies C = 0$

Thus, the constant $C$ is zero.

Step 4: Final velocity equation

The velocity as a function of time is:

$v(t) = P t - \frac{Q t^2}{2}$

This equation describes the car's velocity as a function of time given the time-varying acceleration.

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Do you have any questions or would you like further details?

Here are 5 related questions:

1. How would the velocity change if $P$ were larger?
2. What happens to the velocity when $t = \frac{2P}{Q}$?
3. How does this velocity function relate to the displacement of the car?
4. What would the acceleration look like graphically as a function of time?
5. How would the equation change if $Q$ were negative?

Tip: When integrating, always remember to include the constant of integration and use initial conditions to determine its value.