Problem 73 - Fundamentals of Physics

Question

At the instant the traffic light turns green, an automobile starts with a constant acceleration \(a\) of \(2.2 \mathrm{~m} / \mathrm{s}^{2}\). At the same instant a truck, traveling with a constant speed of \(9.5 \mathrm{~m} / \mathrm{s}\), overtakes and passes the automobile. (a) How far beyond the traffic signal will the automobile overtake the truck? (b) How fast will the automobile be traveling at that instant?

Step-by-Step Solution

Answer

The automobile overtakes the truck at approximately 82.08 meters, traveling at 19.01 m/s.

Step 1: Understand the Problem

We need to find two things: (a) the distance where the automobile overtakes the truck and (b) the speed of the automobile at that point. The automobile accelerates from rest while the truck travels at a constant speed of 9.5 m/s.

Step 2: Define Variables

Let \( d \) be the distance at which the automobile overtakes the truck.Let \( t \) be the time it takes for this to happen.The acceleration \( a \) of the car is 2.2 m/s².The speed of the truck \( v_{t} \) is 9.5 m/s.

Step 3: Set Up Equations for the Distance

For the truck moving at a constant speed, the distance equation is:\[ d = v_{t} \, t \]For the automobile, starting from rest, the distance equation is:\[ d = \frac{1}{2} \, a \, t^{2} \]

Step 4: Equate Distances to Find Time

Because the automobile overtakes the truck at distance \( d \), equate the distance equations:\[ \frac{1}{2} \, a \, t^{2} = v_{t} \, t \]Substitute the known values: \( a = 2.2 \) m/s² and \( v_{t} = 9.5 \) m/s:\[ \frac{1}{2} \, (2.2) \, t^{2} = 9.5 \, t \]Simplifying gives:\[ 1.1 \, t^{2} = 9.5 \, t \]Solving for \( t \):\[ t = \frac{9.5}{1.1} \approx 8.64 \text{ seconds} \]

Step 5: Calculate the Distance

Substitute \( t \approx 8.64 \) into the distance formula for the truck:\[ d = 9.5 \, t = 9.5 \times 8.64 \approx 82.08 \text{ meters} \]

Step 6: Calculate the Speed of the Automobile

Using the formula for velocity with constant acceleration, \( v = a \, t \):Substitute the known values \( a = 2.2 \) m/s² and \( t \approx 8.64 \) seconds:\[ v = 2.2 \times 8.64 \approx 19.01 \text{ m/s} \]

Key Concepts

Constant Acceleration

In the world of kinematics, constant acceleration refers to a scenario where an object's acceleration remains the same over time. This means that its velocity changes linearly with time.

• In the given exercise, the automobile has a constant acceleration of 2.2 m/s². This implies that every second, the car's velocity increases by 2.2 m/s.
• This is the primary influence on the car's motion as it starts from rest, meaning an initial velocity of 0 m/s.

The motion of objects under constant acceleration is described by the equation: \[ d = rac{1}{2} a t^2 + v_i t\]where:

• \( d \) is the distance traveled,
• \( a \) is the acceleration,
• \( t \) is the time,
• \( v_i \) is the initial velocity, which in this case is 0.

This basis allows us to predict the future position and speed of the car, assuming it keeps accelerating at this known rate. It is a foundational concept for solving problems involving accelerating objects, like in our exercise.

Uniform Velocity

Uniform velocity describes a situation where an object moves at a consistent speed in a straight line. Its speed remains constant, meaning that there is no acceleration involved.

• In the exercise, the truck is moving with a uniform velocity of 9.5 m/s.
• This implies that regardless of the time t, the truck covers the same distance every second.

The formula for uniform velocity is simply:\[ d = v \, t \]where:

• \( d \) is the distance traveled,
• \( v \) is the velocity,
• \( t \) is the time.

This straightforward relationship makes calculations with constant speeds quite intuitive. In our problem, the truck's consistent velocity allows us to easily determine the distance it covers over a given period by multiplying its velocity by time.

Distance-Time Relationship

The distance-time relationship is a fundamental concept in kinematics. It explains how the distance traveled by an object relates to the time it has been moving.For different types of motion, this relationship is described by various equations:

• For constant acceleration, the distance is given by:\( d = rac{1}{2} a t^2\)This indicates that the distance increases with the square of time.
• For uniform velocity, the distance is calculated with:\( d = v t \)This shows a direct proportionality between distance and time.

In our exercise:

• The car's distance increases quadratically because of constant acceleration from rest.
• The truck's distance increases linearly due to its uniform velocity.

By comparing these distances over time, the exercise asks us when they become equal—when the car overtakes the truck.

Velocity-Time Relationship

The velocity-time relationship characterizes how the velocity of an object changes over time, a principal concept in kinematics.With constant acceleration, the velocity of an object is described by:\[ v = v_i + a t \]where:

• \( v \) is the final velocity,
• \( v_i \) is the initial velocity,
• \( a \) is the acceleration,
• \( t \) is the time.

For our exercise:

• The automobile starts from rest, so \( v_i = 0 \), making the velocity-time equation simply \( v = a t \).
• After 8.64 seconds, the car achieves a velocity of 19.01 m/s.
• The truck, on the other hand, maintains a constant velocity of 9.5 m/s, not influenced by time since its acceleration is zero.

Understanding these relationships helps solve for various aspects of objects' motion, determining both velocity at a given point in time and how speed changes with time for objects under acceleration.