Understanding Acceleration in a Decelerating Car Scenario: A Practical Example and Analysis

Have you ever noticed how cars come to a stop, often leaving witnesses with the impression that they are acting as if they were accelerating in the opposite direction? This phenomenon can be mathematically analyzed using basic principles of physics, particularly Newton's laws and kinematic equations. Let's explore a real-world example to understand the concept of acceleration and deceleration in a car.

A Real-World Example

Consider a car moving along a straight highway with a speed of 72 km/h. This car is brought to a stop within a distance of 200 meters. We want to calculate the deceleration of the car. To solve this, we can use the following kinematic equation:

v^2 u^2 - 2as

Where:

v final velocity (0 m/s, since the car comes to a stop) u initial velocity in m/s a acceleration in m/s2 s distance 200 m

Let's start by converting the initial speed from km/h to m/s:

u 72 km/h * (1000 m / 1 km) * (1 h / 3600 s) 20 m/s

Substitute the known values into the equation:

0 (202) - (2a * 200)

This simplifies to:

0 400 - 400a

Rearranging the equation:

400a -400

Solving for a we get:

a -1 m/s2

Therefore, the acceleration of the car is -1 m/s2, indicating it is decelerating.

Analysis and Definitions

Definition of Acceleration: In physics, acceleration (a) is defined as the rate of change of velocity (v) with respect to time (t), represented mathematically as:

a Delta;v / Delta;t

Given Values:

Initial Velocity (vi) 20 m/s Final Velocity (vf) 0 m/s Distance (s) 200 m

Delta;v: The change in velocity

Delta;v vf - vi 0 - 20 -20 m/s

Delta;t: The time taken to stop

First, let's calculate the average velocity (vavg):

vavg (vi vf) / 2 (20 0) / 2 10 m/s

Then, the time taken to stop (tf):

tf s / vavg 200 m / 10 m/s 20 s

Delta;t:

Delta;t tf - ti 20 - 0 20 s

Now we can calculate the acceleration:

a Delta;v / Delta;t -20 m/s / 20 s -1 m/s2

As expected, the car is decelerating, and the negative sign indicates the direction of acceleration.

Visualizing the Process

Let's visualize the equations used to solve the problem:

d Vot 1/2at2

Vf2 - Vi2 2ad

02 - 202 2a * 200

-400 400a

a -1 m/s2

Another way to look at it:

Average velocity is 1/2 of the initial velocity if uniform deceleration, which is 10 m/s. So, the time to stop is 200 m / 10 m/s 20 s. The deceleration according to V at is just 1 m/s2.

The given data:

Initial Velocity of the car (u) 72 km/h 20 m/s

Final Velocity (v) 0 m/s

Distance covered (s) 200 m

Applying the third equation of motion:

v2 u2 - 2as

02 202 - 2a * 200

0 400 - 400a

400a -400

a -1 m/s2

Hence, the acceleration of the car is -1 m/s2, or the retardation is 1 m/s2.