Question

A car accelerates from a standing start to 60.0 mph in 4.20 s. Assume that its acceleration is constant. a) What is the acceleration? b) How far does the car travel?

Short answer

Question: A car accelerates from a standing start (initial velocity = 0) to 60.0 mph in 4.20 seconds. Calculate (a) the acceleration of the car and (b) the distance it travels during these 4.20 seconds. Answer: (a) The acceleration of the car is 6.3862 m/s^2. (b) The car travels 56.367 meters during the 4.20 seconds.

Step-by-step solution

(a) Calculate the acceleration

Firstly, we need to convert the final velocity from mph to m/s. 60.0 mph = 60.0 * 1609.34 / 3600 = 26.8224 m/s. Now, we have the initial velocity (u) = 0 m/s, final velocity (v) = 26.8224 m/s, and time (t) = 4.20 s. We can use the equation: v = u + at to find the acceleration 'a'. Rearranging the equation to solve for 'a', we get: a = (v - u) / t a = (26.8224 - 0) / 4.20 a = 6.3862 m/s^2 The acceleration of the car is 6.3862 m/s^2.

(b) Calculate the distance traveled

Having found the acceleration, we can now determine the distance traveled during the 4.20 seconds. We will use the equation: s = ut + (1/2)at^2. Substituting the values in into the equation, we get: s = (0)(4.20) + (1/2)(6.3862)(4.20)^2 s = 0 + 3.1931(17.64) s = 56.367 m The car travels 56.367 meters during the 4.20 seconds.

Key concepts

Kinematic Equations

When analyzing problems involving motion, it's crucial to understand kinematic equations. These equations are the core tools for describing the motion of objects with constant acceleration. Out of the five primary kinematic equations, one commonly used formula is: \( v = u + at \), where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. This equation allows us to calculate the acceleration if we know the initial and final velocities, along with the time taken for the change.

Another vital equation is \( s = ut + \frac{1}{2}at^2 \), which describes the displacement, \( s \) of an object. In this, the first part, \( ut \), accounts for the distance covered at initial velocity, while the second part, \( \frac{1}{2}at^2 \), takes into account the additional distance covered due to acceleration over time. These equations are fundamental for solving most problems where the object's acceleration is uniform.

For a clear understanding and correct application, identifying which kinematic equation to use based on the given information is essential. Each one provides a relationship between different variables such as distance, time, velocity, and acceleration, allowing us to solve for the unknown.

Units Conversion

Units conversion is a necessary step in physics problems to ensure all measurements are in the same system and can be accurately used in formulas. In the provided exercise, the initial step involved converting the speed of the car from miles per hour (mph) to meters per second (m/s) because the standard unit of velocity in the International System of Units (SI) is m/s.

To convert mph to m/s, the conversion factor \( 1609.34 / 3600 \) is used because there are 1609.34 meters in a mile and 3600 seconds in an hour. This precise conversion is important for two reasons: first, to ensure the correct values are plugged into the kinematic equations; and second, to enable consistent comparisons and calculations alongside other physics problems which generally use SI units.

Understanding how to move between different units of measure is a key skill in physics. Multiplying by conversion factors allows for the transformation of a measurement in one unit to its equivalent in another unit. This technique helps in avoiding mistakes that can occur due to the discrepancies in units, especially when dealing with complex calculations involving several steps.

Acceleration and Distance Relationship

The relationship between acceleration and distance is another core concept to comprehend when dealing with constant acceleration scenarios. When an object has a constant acceleration, the distance it covers is directly related to the square of the time it has been accelerating, due to the equation \( s = ut + \frac{1}{2}at^2 \).

This quadratic relationship between time and distance means that as time progresses, the distance covered does not just increase linearly but accelerates. For instance, if the time doubles, the distance covered by the object will be quadrupled, provided the acceleration remains constant. This is a crucial concept when predicting how much distance an object will cover in a given time interval under constant acceleration.

In the example problem, we calculated that with a constant acceleration, a car moving from rest covers a distance proportional to the square of the time it has been moving. By recognizing this relationship, it becomes easier to visualize the object's motion over time and to predict how changes in acceleration or time will affect the distance traveled.