A car is travelling to the left, which is the negative direction. The direction of travel remains the same throughout this problem. The car's initial speed is 27.0 m/s, and during a 5.0 s interval, it changes to a final speed of 29.0 m/s. Find the acceleration ( magnitude and algebraic sign ) and state whether or not the car is decelerating.

Options

A. `-11 m//s^(2)`
B. `-5.8 m//s^(2)`
C. `-0.40 m//s^(2)`
D. `+0.40 m//s^(2)`

Answer

To solve the problem, we need to find the acceleration of the car, which is defined as the change in velocity over time. The car is moving in the negative direction (to the left), and we will denote this direction with a negative sign.

### Step-by-Step Solution:

1. **Identify Initial and Final Velocities:**
- The initial speed of the car (u) is 27.0 m/s to the left, which we denote as:
\[ u = -27.0 \, \text{m/s} \]
- The final speed of the car (v) is 29.0 m/s to the left, denoted as:
\[ v = -29.0 \, \text{m/s} \]

2. **Determine the Time Interval:**
- The time interval (t) during which the speed changes is given as:
\[ t = 5.0 \, \text{s} \]

3. **Use the Formula for Acceleration:**
- The formula for acceleration (a) is given by:
\[ a = \frac{v - u}{t} \]
- Substitute the values of v, u, and t into the formula:
\[ a = \frac{-29.0 \, \text{m/s} - (-27.0 \, \text{m/s})}{5.0 \, \text{s}} \]
- Simplifying this gives:
\[ a = \frac{-29.0 + 27.0}{5.0} = \frac{-2.0}{5.0} = -0.4 \, \text{m/s}^2 \]

4. **Interpret the Result:**
- The acceleration is \(-0.4 \, \text{m/s}^2\). The negative sign indicates that the acceleration is in the same direction as the initial velocity (to the left).
- Since the velocity is increasing (from -27.0 m/s to -29.0 m/s), the car is not decelerating; instead, it is accelerating in the negative direction.

5. **Conclusion:**
- The magnitude of the acceleration is \(0.4 \, \text{m/s}^2\) and the algebraic sign is negative.
- The car is **not** decelerating; it is accelerating in the negative direction.