A car moving with a speed of 12 m/s is subjected to brakes which produces a deceleration of 6 m/s2 The car takes 2s to stop after the application of brakes. What is the distance covered by the car after the application of brakes?
12 m
Calculating Car Stopping Distance with Deceleration
This question asks us to find the distance covered by a car from the moment brakes are applied until it comes to a complete stop. We are given the initial speed of the car, the deceleration produced by the brakes, and the time it takes to stop.
Understanding the Given Information
- Initial speed of the car ($u$) = 12 m/s
- Deceleration produced ($a$) = -6 m/s\textsuperscript{2} (Deceleration is negative acceleration)
- Time taken to stop ($t$) = 2 s
- Final speed of the car ($v$) = 0 m/s (Since the car stops)
We need to calculate the distance covered ($s$) during this braking period.
Choosing the Right Physics Formula
We can use the equations of motion (kinematic equations) for uniformly accelerated motion. We have initial velocity ($u$), final velocity ($v$), acceleration ($a$), and time ($t$). We need to find the distance ($s$).
One suitable equation relating $s$, $u$, $a$, and $t$ is:
\( s = ut + \frac{1}{2}at^2 \)
Another suitable equation relating $v$, $u$, $a$, and $s$ is:
\( v^2 = u^2 + 2as \)
Step-by-Step Calculation using \( s = ut + \frac{1}{2}at^2 \)
Let's use the first equation, as we have all the values ($u$, $t$, $a$).
Substitute the given values into the formula:
\( s = (12 \text{ m/s})(2 \text{ s}) + \frac{1}{2}(-6 \text{ m/s}^2)(2 \text{ s})^2 \)
First, calculate the terms:
- \( ut = 12 \times 2 = 24 \) m
- \( t^2 = 2^2 = 4 \) s\textsuperscript{2}
- \( \frac{1}{2}at^2 = \frac{1}{2} \times (-6) \times 4 \)
- \( \frac{1}{2}at^2 = -3 \times 4 = -12 \) m
Now, add the terms to find $s$:
\( s = 24 \text{ m} + (-12 \text{ m}) \)
\( s = 24 - 12 \)
\( s = 12 \text{ m} \)
Step-by-Step Calculation using \( v^2 = u^2 + 2as \)
Let's also use the second equation to verify the result. We have $v=0$, $u=12$, and $a=-6$. We need to find $s$.
Substitute the values:
\( (0 \text{ m/s})^2 = (12 \text{ m/s})^2 + 2(-6 \text{ m/s}^2)s \)
\( 0 = 144 \text{ m}^2/\text{s}^2 - 12s \text{ m/s}^2 \)
Rearrange the equation to solve for $s$:
\( 12s = 144 \)
\( s = \frac{144}{12} \)
\( s = 12 \text{ m} \)
Both methods give the same result.
Result
The distance covered by the car after the application of brakes until it stops is 12 meters.
| Parameter | Symbol | Value | Unit |
|---|---|---|---|
| Initial Velocity | \(u\) | 12 | m/s |
| Final Velocity | \(v\) | 0 | m/s |
| Deceleration | \(a\) | -6 | m/s\textsuperscript{2} |
| Time | \(t\) | 2 | s |
| Distance | \(s\) | ? | m |
Revision Table: Car Motion Concepts
| Concept | Description | Key Formula |
|---|---|---|
| Velocity | Rate of change of displacement. | \(v = \frac{\Delta x}{\Delta t}\) (average) |
| Acceleration | Rate of change of velocity. | \(a = \frac{\Delta v}{\Delta t}\) (average) |
| Deceleration | Acceleration in the opposite direction of motion, causing speed to decrease. | Negative acceleration (\(-a\)) |
| Displacement/Distance | Change in position or total path length covered. | \(s = ut + \frac{1}{2}at^2\), \(v^2 = u^2 + 2as\), etc. |
Additional Information: Kinematic Equations
The kinematic equations are a set of equations that describe the motion of objects with constant acceleration. They are fundamental tools for solving problems involving linear motion.
The main equations are:
- \( v = u + at \): Relates final velocity, initial velocity, acceleration, and time.
- \( s = ut + \frac{1}{2}at^2 \): Relates displacement, initial velocity, acceleration, and time.
- \( v^2 = u^2 + 2as \): Relates final velocity, initial velocity, acceleration, and displacement.
- \( s = \frac{(u+v)}{2}t \): Relates displacement, initial velocity, final velocity, and time (useful when acceleration is not given or needed).
When solving problems, it's crucial to correctly identify the given quantities and the quantity you need to find, and then select the appropriate equation. Remember that deceleration is simply negative acceleration.
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