At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.20 m/s2. At the same instant a truck, traveling with a constant speed of 20.0 m/s, overtakes and passes the car. How far beyond its starting point is the car traveling when it overtakes the truck?

In this problem, we have a car accelerating from rest and a truck moving at a constant speed. To find the distance beyond its starting point when the car overtakes the truck, we need to set up equations for the positions of both the car and the truck as functions of time, then solve for the point where the car overtakes the truck (i.e., where their positions are equal).

### Step 1: Equations of motion
1. **For the car**:
   The car starts from rest and accelerates with a constant acceleration of \(a = 3.20 \, \text{m/s}^2\). The equation for its position as a function of time is:
   \[
   x_{\text{car}}(t) = \frac{1}{2} a t^2
   \]
   Since the car starts from rest, its initial velocity is zero, so we omit the velocity term.

   Substituting \(a = 3.20 \, \text{m/s}^2\):
   \[
   x_{\text{car}}(t) = \frac{1}{2} \times 3.20 \times t^2 = 1.60 t^2
   \]

2. **For the truck**:
   The truck is moving with a constant speed of \(v = 20.0 \, \text{m/s}\). The equation for its position as a function of time is:
   \[
   x_{\text{truck}}(t) = v t = 20.0 t
   \]

### Step 2: Setting the positions equal
The car overtakes the truck when their positions are equal, so we set \(x_{\text{car}}(t)\) equal to \(x_{\text{truck}}(t)\):
\[
1.60 t^2 = 20.0 t
\]

### Step 3: Solving the equation
First, rearrange the equation:
\[
1.60 t^2 - 20.0 t = 0
\]
Factor out \(t\):
\[
t(1.60 t - 20.0) = 0
\]
This gives two solutions:
1. \(t = 0\) (which is the initial time, not the point when the car overtakes the truck),
2. \(1.60 t = 20.0\), or:
   \[
   t = \frac{20.0}{1.60} = 12.5 \, \text{seconds}
   \]

### Step 4: Finding the distance
Now that we know the car overtakes the truck after \(t = 12.5 \, \text{seconds}\), we can find how far the car has traveled by substituting this time into the car's position equation:
\[
x_{\text{car}}(12.5) = 1.60 \times (12.5)^2
\]
\[
x_{\text{car}}(12.5) = 1.60 \times 156.25 = 250.0 \, \text{meters}
\]

### Final Answer:
The car overtakes the truck 250.0 meters beyond its starting point.

Would you like a detailed breakdown of any step or further explanations?

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Here are 5 related questions:
1. How would the result change if the truck was accelerating instead of moving at constant speed?
2. What if the car had a different acceleration, say 5.0 m/s², how would that affect the time it overtakes the truck?
3. What is the truck's position after 12.5 seconds, and how does it relate to the car's position?
4. How would this problem change if there was a delay before the car started accelerating?
5. How can you calculate the velocity of the car when it overtakes the truck?

**Tip:** When solving motion problems, always clearly define your reference point and initial conditions to avoid confusion.

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.20 m/s². At the same instant a truck, traveling with a constant speed of 20.0 m/s, overtakes and passes the car. How far beyond its starting point is the car traveling when it overtakes the truck?

In this kinematics problem, a car accelerates from rest while a truck moves at a constant speed. Find the distance beyond the starting point where the car overtakes the truck, using motion equations. Detailed step-by-step solution provided. Suitable for students in Grades 10-12.

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