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 ms^(-2)`, At the some instant, a truck travelling with a constant speed of `20.0 ms^(-1)`, overtakes and passes the car.
a. At what distance from its starting point does the car overtake the truck?
b. Calculate the speed of the car when it overtakes the truck.
c. Sketch an `x-t` graph of themotion of both vehicles.
Take `x-0` at the intersection.
d. Sketch a `v_(x)-t` graph of the motion of both vehicles.

Explanation

Let's solve the problem step by step.

### Given Data:
- Acceleration of the car, \( a = 3.20 \, \text{m/s}^2 \)
- Initial velocity of the car, \( u_{\text{car}} = 0 \, \text{m/s} \)
- Speed of the truck, \( v_{\text{truck}} = 20.0 \, \text{m/s} \)
- Initial velocity of the truck, \( u_{\text{truck}} = 20.0 \, \text{m/s} \)
- Initial position of both vehicles, \( x_0 = 0 \)

### Part (a): Distance from the starting point where the car overtakes the truck

1. **Distance traveled by the truck in time \( t \)**:
\[ x_{\text{truck}} = v_{\text{truck}} \cdot t = 20.0 \cdot t \quad \text{(Equation 1)} \]

2. **Distance traveled by the car in time \( t \)**:
\[ x_{\text{car}} = u_{\text{car}} \cdot t + \frac{1}{2} a t^2 = 0 \cdot t + \frac{1}{2} \cdot 3.20 \cdot t^2 = 1.6 t^2 \quad \text{(Equation 2)} \]

3. **Setting the distances equal** (since the car overtakes the truck):
\[ 20.0 t = 1.6 t^2 \]

4. **Rearranging the equation**:
\[ 1.6 t^2 - 20.0 t = 0 \]
\[ t(1.6 t - 20.0) = 0 \]

5. **Solving for \( t \)**:
- \( t = 0 \) (initial time, not relevant)
- \( 1.6 t - 20.0 = 0 \)
\[ t = \frac{20.0}{1.6} = 12.5 \, \text{s} \]

6. **Finding the distance where the car overtakes the truck**:
Substitute \( t = 12.5 \) s into Equation 1:
\[ x_{\text{truck}} = 20.0 \cdot 12.5 = 250 \, \text{m} \]

### Part (b): Speed of the car when it overtakes the truck

1. **Using the formula for final velocity**:
\[ v = u + at \]
\[ v_{\text{car}} = 0 + 3.20 \cdot 12.5 \]
\[ v_{\text{car}} = 40.0 \, \text{m/s} \]

### Part (c): Sketch an \( x-t \) graph of the motion of both vehicles

- **Truck**: The graph will be a straight line with a slope of \( 20 \, \text{m/s} \).
- **Car**: The graph will be a parabola starting from the origin, as it accelerates from rest.

### Part (d): Sketch a \( v-t \) graph of the motion of both vehicles

- **Truck**: The graph will be a horizontal line at \( 20 \, \text{m/s} \).
- **Car**: The graph will be a straight line starting from the origin and increasing to \( 40 \, \text{m/s} \) at \( t = 12.5 \, \text{s} \).

### Summary of Results:
- **Distance where the car overtakes the truck**: \( 250 \, \text{m} \)
- **Speed of the car when it overtakes the truck**: \( 40 \, \text{m/s} \)