A car is driving along a straight, horizontal road, as shown in Figure 1. Starting from rest at time t = 0, the car
moves with an acceleration a that varies with time t according to the equation a(t) = P - Qt, where P and Q
are positive constants and the positive direction is taken to be toward the right.
(a)
i. The dots in Figure 2 represent the car at two instants in time, t1 =
20 and t2 =
26. On each dot, draw an
arrow to represent the direction, if any, of the car's acceleration at that time. If the acceleration is zero at either
time, write 'a = 0' next to the appropriate dot.

To solve this problem, we need to analyze the acceleration function \( a(t) = P - Qt \), where \( P \) and \( Q \) are positive constants.

### Step 1: Understanding the acceleration function
- The acceleration \( a(t) = P - Qt \) starts with a positive value \( P \) at \( t = 0 \), since \( Q \) and \( t \) are both zero initially.
- As time \( t \) increases, the term \( Qt \) increases, which causes the acceleration to decrease.
- Eventually, the acceleration will reach zero when \( P = Qt \), or at \( t = \frac{P}{Q} \).
- For times \( t > \frac{P}{Q} \), the acceleration becomes negative, indicating deceleration in the positive direction.

### Step 2: Analyzing the acceleration at \( t_1 = 20 \) and \( t_2 = 26 \)
- To determine whether the acceleration is positive, negative, or zero at these times, we need to evaluate the acceleration function at these specific time values.
  
  **At \( t_1 = 20 \):**
  \[
  a(20) = P - Q \cdot 20
  \]
  If \( P > 20Q \), the acceleration will still be positive at \( t = 20 \), indicating the car is still accelerating in the positive direction.
  If \( P < 20Q \), the acceleration will be negative at \( t = 20 \), indicating the car is decelerating in the positive direction.

  **At \( t_2 = 26 \):**
  \[
  a(26) = P - Q \cdot 26
  \]
  Similarly, if \( P > 26Q \), the acceleration is still positive, but if \( P < 26Q \), the acceleration is negative.

### Step 3: Drawing the arrows on the dots in Figure 2
- For \( t_1 = 20 \), if \( a(20) > 0 \), draw the arrow pointing to the right to indicate positive acceleration. If \( a(20) < 0 \), draw the arrow to the left to indicate negative acceleration.
- For \( t_2 = 26 \), if \( a(26) > 0 \), draw the arrow to the right; if \( a(26) < 0 \), draw it to the left.
- If the acceleration equals zero at any time, write "a = 0" next to the dot.

Would you like me to elaborate on the conditions for determining the sign of acceleration, or is there a specific calculation you have in mind?

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**Follow-up Questions:**
1. What would happen if \( P = 20Q \) or \( P = 26Q \)? How would it change the diagram?
2. How does the velocity change when the acceleration becomes negative?
3. How can you determine the time when the car comes to a stop if the velocity is initially increasing?
4. What would the graph of acceleration versus time look like for this function?
5. How does the choice of positive direction affect the interpretation of the problem?

**Tip:** When acceleration is given as a function of time, integrate it with respect to time to find the velocity function, and integrate again to find the position function.

A car is driving along a straight, horizontal road, as shown in Figure 1. Starting from rest at time t = 0, the car moves with an acceleration a that varies with time t according to the equation a(t) = P - Qt, where P and Q are positive constants and the positive direction is taken to be toward the right. (a) i. The dots in Figure 2 represent the car at two instants in time, t1 = 20 and t2 = 26. On each dot, draw an arrow to represent the direction, if any, of the car's acceleration at that time. If the acceleration is zero at either time, write 'a = 0' next to the appropriate dot.

Learn how to analyze the motion of a car with a variable acceleration function a(t) = P - Qt. Understand the conditions where acceleration changes direction, becomes zero, or indicates deceleration. This problem involves key kinematics concepts and is suitable for students in Grades 10-12.

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