To determine where the car reaches the truck, we need to find the point where both vehicles have traveled the same distance from the starting point.

Let:

• $a = 5.50 \, \text{m/s}^2$ be the car's acceleration.
• $v_t = 18.0 \, \text{m/s}$ be the truck's constant speed.
• $s$ be the distance from the starting point where they meet.
• $t$ be the time it takes for the car to reach the truck.

Step 1: Set up equations for distance traveled by each vehicle

1. For the car (starting from rest with constant acceleration): $s_{\text{car}} = \frac{1}{2} a t^2$ Substituting $a = 5.50 \, \text{m/s}^2$: $s_{\text{car}} = \frac{1}{2} \cdot 5.50 \, t^2 = 2.75 \, t^2$

2. For the truck (moving at constant speed): $s_{\text{truck}} = v_t \cdot t$ Substituting $v_t = 18.0 \, \text{m/s}$: $s_{\text{truck}} = 18.0 \, t$

Step 2: Set the distances equal to each other

Since they meet at the same point, the distances are equal: $2.75 \, t^2 = 18.0 \, t$

Step 3: Solve for $t$

Rearrange the equation: $2.75 \, t^2 - 18.0 \, t = 0$ Factor out $t$: $t(2.75 \, t - 18.0) = 0$ This gives two solutions:

1. $t = 0$ (the starting point)
2. $t = \frac{18.0}{2.75}$

Calculate $t$: $t = \frac{18.0}{2.75} \approx 6.55 \, \text{seconds}$

Step 4: Find the distance $s$ where they meet

Now substitute $t = 6.55$ seconds into either distance equation. Using the truck’s equation: $s = 18.0 \times 6.55 \approx 117.9 \, \text{meters}$

Answer

The car reaches the truck approximately 117.9 meters from the starting point.

Would you like a breakdown of the steps, or have any questions?

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1. How would the answer change if the truck’s speed were different?
2. What if the car had an initial speed instead of starting from rest?
3. How does the result differ if the car’s acceleration changes halfway?
4. What would happen if the truck also accelerated?
5. How long would it take if the car’s acceleration was only half of what it is?

Tip: When two objects start at the same time with different motions, set their distance equations equal to find where they meet.