To solve the problem, we need to analyze the motion of the car given the acceleration function $a(t) = P - Qt$, where $P$ and $Q$ are constants, and $t$ is the time.

Step 1: Find the Velocity Function

Acceleration is the derivative of velocity with respect to time, so to find the velocity, we integrate the acceleration function with respect to time:

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

$v(t) = Pt - \frac{Qt^2}{2} + C$

Since the car starts from rest at $t = 0$, the initial velocity $v(0)$ is 0. Therefore:

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

So, the velocity function is:

$v(t) = Pt - \frac{Qt^2}{2}$

Step 2: Find the Position Function

Velocity is the derivative of position with respect to time, so to find the position function $x(t)$, we integrate the velocity function:

$x(t) = \int v(t) \, dt = \int \left(Pt - \frac{Qt^2}{2}\right) dt$

$x(t) = \frac{Pt^2}{2} - \frac{Qt^3}{6} + D$

Again, since the car starts from rest at $t = 0$ from the initial position $x(0) = 0$:

$x(0) = 0 = \frac{P(0)^2}{2} - \frac{Q(0)^3}{6} + D \implies D = 0$

So, the position function is:

$x(t) = \frac{Pt^2}{2} - \frac{Qt^3}{6}$

Summary of Results

• Acceleration: $a(t) = P - Qt$
• Velocity: $v(t) = Pt - \frac{Qt^2}{2}$
• Position: $x(t) = \frac{Pt^2}{2} - \frac{Qt^3}{6}$

Would you like any further details on this solution, or do you have any specific questions?

Related Questions

1. How would the motion change if the acceleration was constant instead of time-dependent?
2. What happens to the velocity and position functions as time $t$ increases?
3. How would you find the time when the car comes to a stop?
4. What physical meaning can be derived if $P$ and $Q$ are equal?
5. How would you analyze the motion if $Q$ were negative?

Tip

When dealing with kinematic equations, always start by identifying the basic relationships between acceleration, velocity, and position, and use integration and initial conditions to find the specific functions.