Problem 20 - Applied Mathematics

Question

A car is brought to rest by the action of its brakes which are assumed to exert a constant force on the car. Find the distance moved by the car before coming to rest. (a) The mass of the car is \(750 \mathrm{~kg}\). (b) The initial velocity of the car is \(40 \mathrm{~ms}^{-1}\). (c) The time taken is 5 seconds.

Step-by-Step Solution

Answer

The car moved 100 meters before coming to rest.

Step 1: Identify the given values

The problem provides the following values:- Mass of the car, \( m = 750 \, \text{kg} \)- Initial velocity, \( u = 40 \, \text{m/s} \)- Time taken to stop, \( t = 5 \, \text{s} \)

Step 2: Understand the final condition

Since the car comes to rest, the final velocity \( v \) of the car is \( 0 \, \text{m/s} \).

Step 3: Calculate the acceleration

Use the equation of motion: \[ v = u + a t \]Substituting the given values: \[ 0 = 40 + a \times 5 \]Solving for \( a \): \[ a = -\frac{40}{5} = -8 \, \text{m/s}^2 \](Note: the negative sign indicates deceleration.)

Step 4: Calculate the distance moved

Use the equation of motion: \[ v^2 = u^2 + 2 a s \]Substituting the known values: \[ 0 = 40^2 + 2 (-8) s \]Solving for \( s \): \[ 0 = 1600 - 16s \]\[ 16s = 1600 \]\[ s = \frac{1600}{16} = 100 \, \text{m} \]

Key Concepts

Constant Force

A constant force means that the force applied on the car remains steady over the duration of the event. In this problem, the car's brakes exert a constant force, causing it to decelerate and eventually come to a stop. This steady application simplifies our calculations, as we do not have to consider changing forces. Imagine pushing a shopping cart with the same amount of force until it stops; you maintain the same effort throughout.

Equations of Motion

Equations of motion help us calculate entities like displacement and velocity under constant acceleration or deceleration. In our problem, we use these equations:
- First Equation of Motion: \( v = u + at \)
This equation relates final velocity (v), initial velocity (u), acceleration (a), and time (t). For our car:
\( 0 = 40 + a \times 5 \)
Solving for the acceleration (a), we get:
\[ a = -\frac{40}{5} = -8 \text{ m/s}^2 \]
- Second Equation of Motion: \( v^2 = u^2 + 2as \)
This equation lets us calculate the displacement (s) of the car given the initial velocity (u), final velocity (v), and acceleration (a):
\( 0 = 40^2 + 2(-8)s \)
Solving for displacement:\( s = \frac{1600}{16} = 100 \text{ m} \).
These equations are fundamental for calculating motion in physics. You just plug in the known values to find the unknowns.

Distance Calculation

Calculating the distance (displacement) moved by the car before it stops involves understanding the relationship between velocity, acceleration, and distance. We previously derived the acceleration to be \(-8 \text{ m/s}^2\) using the first equation of motion. Now, let's use that:
We know the initial velocity (u) is 40 m/s, the final velocity (v) is 0 m/s, and acceleration (a) is -8 m/s².
Using the second equation of motion: \( v^2 = u^2 + 2as \)
Plugging in the values:
\( 0 = 40^2 + 2(-8)s \)
Simplifying, we get:
\( 1600 = 16s \)
Solving for s (distance):
\( s = \frac{1600}{16} = 100 \text{ m} \)
Our car travels 100 meters before coming to a stop. This method provides a structured approach to solving distance problems in physics.

Deceleration

Deceleration is simply negative acceleration, indicating a reduction in velocity. In the exercise, the car's velocity decreases due to the force from the brakes. The negative sign in our acceleration calculation (\(-8 \text{ m/s}^2\)) signifies deceleration. This means the car's speed is reducing by 8 meters per second every second. Understanding deceleration is key in real-life scenarios like braking in a car, which involves reducing speed effectively and safely.
- **Notice**: Deceleration works opposite to the direction of motion.
Conceptually, whenever you see a negative acceleration value like \(-8 \text{ m/s}^2\), know it's about slowing down, not speeding up. For such problems, always verify if the force reduces the velocity over time, indicating deceleration.
These explanations should clear up how to approach deceleration and distance calculations in physics homework effectively.