at instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a constant acceleration of 2.5 m/s^2. At the same time, a truck, travelling with the constant velocity of 15 m/s overtakes and passes the automobile. How fast is the automobile travelling when it overtakes the truck?

Answer with GRESA method

The GRESA method is a systematic approach used to solve physics problems, standing for **Given, Required, Equation, Solution, Answer**. Let's solve the problem using this approach:

### **G** - Given:
- The automobile has an initial velocity of 0 m/s and accelerates with a constant acceleration of \( a = 2.5 \, \text{m/s}^2 \).
- The truck moves with a constant velocity of \( v_{\text{truck}} = 15 \, \text{m/s} \).
- The two vehicles start moving at the same time, with the truck overtaking the automobile initially.

### **R** - Required:
We need to find the velocity of the automobile, \( v_{\text{auto}} \), when it overtakes the truck.

### **E** - Equation:
We will use the equations of motion to solve this. 

For the **truck**, since it is moving with constant velocity:
\[
x_{\text{truck}} = v_{\text{truck}} \cdot t
\]
For the **automobile**, which is accelerating from rest:
\[
x_{\text{auto}} = \frac{1}{2} a t^2
\]
At the moment when the automobile overtakes the truck, the distance traveled by both the truck and the automobile will be the same:
\[
x_{\text{auto}} = x_{\text{truck}}
\]
Thus, we can set the two equations equal to each other to solve for the time \( t \) when the overtaking happens:
\[
\frac{1}{2} a t^2 = v_{\text{truck}} \cdot t
\]
This equation will allow us to solve for \( t \), and then we can find the velocity of the automobile at that time using the equation for velocity under constant acceleration:
\[
v_{\text{auto}} = a \cdot t
\]

### **S** - Solution:

1. **Set the distances equal**:
   \[
   \frac{1}{2} \cdot 2.5 \cdot t^2 = 15 \cdot t
   \]
2. Simplify the equation:
   \[
   1.25 t^2 = 15 t
   \]
3. Cancel out \( t \) (assuming \( t \neq 0 \)):
   \[
   1.25 t = 15
   \]
4. Solve for \( t \):
   \[
   t = \frac{15}{1.25} = 12 \, \text{seconds}
   \]

5. Now that we know the time \( t \) when the automobile overtakes the truck, we can find the velocity of the automobile at that instant:
   \[
   v_{\text{auto}} = a \cdot t = 2.5 \cdot 12 = 30 \, \text{m/s}
   \]

### **A** - Answer:
The automobile is traveling at \( 30 \, \text{m/s} \) when it overtakes the truck.

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Do you want further details or have any questions?

Here are 5 related questions:
1. How long does it take for the automobile to overtake the truck?
2. What is the distance traveled by both vehicles when the automobile overtakes the truck?
3. What would happen if the truck’s velocity was slower?
4. How would the problem change if the automobile had an initial velocity?
5. What is the significance of constant acceleration in this context?

**Tip:** Always double-check units and make sure they are consistent throughout your calculations.

At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a constant acceleration of 2.5 m/s^2. At the same time, a truck, travelling with the constant velocity of 15 m/s, overtakes and passes the automobile. How fast is the automobile travelling when it overtakes the truck?

This problem explores how an automobile starting with constant acceleration of 2.5 m/s^2 overtakes a truck moving at 15 m/s. Learn the detailed steps to find the automobile's velocity when it overtakes the truck using equations of motion. Suitable for Grades 10-12.

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