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This problem involves both constant acceleration (for the car) and constant velocity (for the truck). We need to find out how far the car travels before it overtakes the truck and the speed at that point.

### Given:
- The car starts from rest with an acceleration of \( a = 6.0 \, \text{m/s}^2 \).
- The truck moves with a constant velocity of \( v_{\text{truck}} = 21 \, \text{m/s} \).
- The car overtakes the truck after traveling a distance \( d \), at which point both the car and truck have traveled for the same amount of time, \( t \).

### Part (a): How far will the car travel before it overtakes the truck?

#### Step 1: Equation for the truck's position
Since the truck moves with constant velocity, its position \( x_{\text{truck}} \) after time \( t \) is given by:
\[
x_{\text{truck}} = v_{\text{truck}} \cdot t = 21t
\]

#### Step 2: Equation for the car's position
For the car, starting from rest and accelerating at a constant rate, its position \( x_{\text{car}} \) after time \( t \) is given by the kinematic equation:
\[
x_{\text{car}} = \frac{1}{2} a t^2 = \frac{1}{2} (6.0) t^2 = 3t^2
\]

#### Step 3: Set the positions equal
At the point the car overtakes the truck, both have traveled the same distance. So, we set their positions equal:
\[
3t^2 = 21t
\]
This simplifies to:
\[
t(3t - 21) = 0
\]
Thus, \( t = 0 \) (initially they are together) or \( t = 7 \) seconds (the time when the car overtakes the truck).

#### Step 4: Find the distance traveled
Now, we can substitute \( t = 7 \) seconds into the equation for the car's position to find the distance:
\[
x_{\text{car}} = 3(7^2) = 3(49) = 147 \, \text{m}
\]
So, the car travels **147 meters** before overtaking the truck.

---

### Part (b): How fast will the car be traveling when it overtakes the truck?

To find the car's speed when it overtakes the truck, we use the equation for velocity under constant acceleration:
\[
v_{\text{car}} = a \cdot t = 6.0 \cdot 7 = 42 \, \text{m/s}
\]
So, the car will be traveling at **42 m/s** when it overtakes the truck.

---

Would you like more details or have any questions?

Here are 5 related questions for further practice:
1. How long would it take the car to reach 63 m/s if it continues to accelerate at the same rate?
2. What is the relative velocity between the car and the truck when the car overtakes the truck?
3. How far apart will the truck and car be 5 seconds after the light turns green?
4. If the truck's velocity were increased to 30 m/s, how long would it take for the car to overtake it?
5. If the car had an initial velocity of 5 m/s instead of starting from rest, how would the time to overtake the truck change?

**Tip**: In motion problems with constant acceleration, it's useful to remember the basic kinematic equations and apply them step by step.

Just as a traffic light turns green, a waiting car starts off with a constant acceleration of 6.0 m/s². At the instant the car begins to accelerate, a truck with a constant velocity of 21 m/s passes in the next lane. (a) How far will the car travel before it overtakes the truck? (b) How fast will the car be traveling when it overtakes the truck?

Solve a kinematics problem involving a car accelerating from rest and overtaking a truck moving at constant speed. Learn how to apply the equations of motion to find the distance traveled by the car and its speed at the point of overtaking. Suitable for students in Grades 9-12.

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