In the opening paragraphs, we will provide a quick answer to the question posed in the title using the given information. Then we will expand on the details and explain the full calculation step-by-step.

The quick answer is that the acceleration of the car is 4 m/s^{2}. This is determined using the kinematic equations of motion with the given values of initial velocity (0 m/s), final velocity (20 m/s) and time taken (5 s). The acceleration is the change in velocity (20 – 0 = 20 m/s) divided by the time taken (5 s), giving 4 m/s^{2}.

## Explanation of Parameters

To fully understand how to calculate the acceleration, we first need to define the parameters involved:

- Initial velocity (v
_{0}) – This is the starting velocity of the object, which is 0 m/s since the car starts from rest. - Final velocity (v
_{f}) – This is the final velocity the object reaches, which is 20 m/s as stated in the question. - Time taken (t) – The time taken for the car to reach the final velocity from rest, which is 5 s.
- Acceleration (a) – This is the rate of change of velocity with time. It represents how quickly (or slowly) the object speeds up or slows down.

These parameters are related through the kinematic equations of motion. The main kinematic equation we will use here is:

a = (v_{f} – v_{0}) / t

Where a is acceleration, v_{f} is final velocity, v_{0} is initial velocity and t is time.

## Step-by-Step Working

Using the kinematic equation above, we can calculate the acceleration step-by-step:

- Initial velocity v
_{0}= 0 m/s (given) - Final velocity v
_{f}= 20 m/s (given) - Time t = 5 s (given)
- Acceleration a = ? (to be calculated)
- Plugging the values into the equation:

a = (v_{f}– v_{0}) / t

a = (20 – 0) m/s / 5 s

a = 20 m/s / 5 s

a = 4 m/s^{2}

Therefore, the acceleration of the car is 4 m/s^{2}.

## Interpreting the Result

The result of 4 m/s^{2} means that for every second that passes, the velocity of the car increases by 4 m/s. This is a moderate acceleration, faster than everyday driving but slower than high-performance sports cars.

Some key points about the calculated acceleration:

- It is positive, indicating the car is speeding up from rest.
- A value of 4 m/s
^{2}means the car reaches 20 m/s (45 mph) in just 5 seconds after starting from 0 m/s. - The rate of change of velocity is constant, since the acceleration is the same over the time interval.
- Over 5 seconds, the velocity changes from 0 to 20 m/s due to the steady increase of 4 m/s per second.

## Relating Acceleration to Everyday Motion

To get a better sense of this acceleration value, we can compare it to some accelerations seen in everyday life and physics:

Motion | Typical Acceleration |
---|---|

Car accelerating slowly from rest | 1.5 m/s^{2} |

Car accelerating moderately from rest | 4 m/s^{2} (matches calculated value) |

High performance sports car accelerating rapidly | 12 m/s^{2} |

Object falling due to gravity near Earth’s surface | 9.8 m/s^{2} |

This shows that an acceleration of 4 m/s^{2} is reasonable for a car accelerating at a moderate pace from rest. It is faster than normal driving but well below the maximum acceleration of sports cars and free falling objects.

## Calculating Additional Parameters

Using the acceleration, we can also calculate some additional parameters like the distance traveled by the car during the 5 seconds:

- Acceleration a = 4 m/s
^{2}(from previous working) - Time t = 5 s (given)
- Initial velocity v
_{0}= 0 m/s (given) - Use equation v
_{f}= v_{0}+ at

v_{f}= 0 + (4)(5)

v_{f}= 20 m/s (matches given value) - Use equation d = v
_{0}t + 1/2at^{2}

d = (0)(5) + 0.5(4)(5)^{2}

d = 0 + 50

d = 50 m

Where d is the distance traveled. Therefore, in the 5 seconds the car covers a distance of 50 m while accelerating from rest to 20 m/s.

## Scenario Variations

Now let’s consider a few variations on the original scenario and see how the acceleration changes:

### Variation 1: Faster Final Velocity

If the final velocity reached was 30 m/s instead of 20 m/s, with the same 5 s time:

- v
_{0}= 0 m/s - v
_{f}= 30 m/s - t = 5 s
- a = (v
_{f}– v_{0}) / t - a = (30 – 0) / 5
- a = 30 / 5
**a = 6 m/s**(increased from 4 m/s^{2}^{2})

The acceleration increases to 6 m/s^{2} if the final velocity is higher while the time remains the same.

### Variation 2: Longer Time

If the car took 10 s instead of 5 s to reach 20 m/s:

- v
_{0}= 0 m/s - v
_{f}= 20 m/s - t = 10 s
- a = (v
_{f}– v_{0}) / t - a = (20 – 0) / 10
- a = 20 / 10
**a = 2 m/s**(decreased from 4 m/s^{2}^{2})

The acceleration decreases to 2 m/s^{2} if the car takes longer to reach the same final velocity.

## Graphical Representation

We can graphically represent the motion of the car accelerating from rest to 20 m/s over 5 seconds. The graph shows how the velocity changes over time at a constant rate of 4 m/s/s:

Key observations:

- Velocity starts at 0 m/s and ramps up linearly over 5 seconds.
- Slope of the line equals the acceleration, 4 m/s
^{2}. - Final velocity of 20 m/s is reached at 5 seconds.

This velocity-time graph visually represents the motion described in the original question with constant acceleration.

## Conclusions

In conclusion, the acceleration of the car starting from rest and reaching 20 m/s in 5 seconds can be calculated using kinematic equations. The acceleration works out to be 4 m/s^{2}.

This indicates a moderately accelerating car, faster than regular driving but slower than sports cars. We calculated additional parameters like distance covered and illustrated graphs. The acceleration varies based on the final velocity reached and the time taken.

Understanding accelerations of moving objects is important in physics and engineering applications. The kinematic equations enable calculating accelerations to gain insights into how objects speed up and slow down.