Question

Two cars are traveling at the same speed, and the drivers hit the brakes at the same time. The deceleration of one car is double that of the other. By what factor does the time required for that car to come to a stop compare with that for the other car?

Short answer

Answer: The time required for the first car to come to a stop is 1/2 (or 0.5 times) that of the other car.

Step-by-step solution

Write the equation for both cars

First, write down the equations for both cars using the third equation of motion: Car 1: 0 = u + a1 * t1 Car 2: 0 = u + a2 * t2 where a1 is the deceleration of Car 1, a2 is the deceleration of Car 2, t1 is the time for Car 1 to stop, and t2 is the time for Car 2 to stop.

Use the given deceleration relationship

We are given that the deceleration of Car 1 is double that of Car 2. We can write this as: a1 = 2 * a2 Now, replace a1 in the equation for Car 1 with this expression: 0 = u + (2 * a2) * t1

Solve for the time relationship

Rearrange each equation to get the time in terms of the deceleration: For Car 1: t1 = -u / (2 * a2) For Car 2: t2 = -u / a2 Now, divide the equation for Car 1 by the equation for Car 2 to get the time factor: t1 / t2 = (-u / (2 * a2)) / (-u / a2)

Simplify the expression and find the time factor

Simplify the expression from Step 3 to get the time factor: t1 / t2 = (-u / (2 * a2)) * (a2 / -u) = 1 / 2 This means that the time required for Car 1 to come to a stop is half of that for Car 2.

Key concepts

Deceleration

When an object slows down, it experiences 'deceleration,' which is a fundamental concept in kinematics and dynamics. In simple terms, deceleration is the reduction in speed or the negative acceleration of an object. It's measured in units of acceleration, which are meters per second squared \(\text{m/s}^2\) in the International System of Units (SI).

Deceleration occurs under various circumstances, such as when a driver applies the brakes in a car. The deceleration value is crucial in determining how quickly a vehicle can come to a stop. The greater the deceleration, the shorter the stopping time and distance. This principle is vividly illustrated in the textbook exercise where Car 1, decelerating at double the rate of Car 2, stops in half the time.

Kinematics

Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause this motion. It involves analyzing velocities, accelerations, and displacements of moving objects.

In kinematics, equations of motion describe the relationships between these quantities. The problem at hand uses one of these equations to relate the initial velocity \(u\), deceleration \(a\), and stopping time \(t\). Understanding these relationships allows students to solve complex problems involving moving objects, such as predicting the motion of two cars with different decelerations.

As the exercise shows, the equation of motion is a powerful tool that directly connects deceleration with the time required to bring an object to rest. This is particularly prevalent in the world of physics problem-solving.

Physics Problem Solving

In physics, problem-solving is an essential skill that involves applying conceptual understanding to concrete situations. Tackling physics problems usually involves several steps: interpreting the problem, identifying the relevant physical principles, applying appropriate formulas or equations, and performing calculations to arrive at a solution.

Improvement in physics problem solving comes from practicing a variety of problems and understanding the underlying principles. For example, in the exercise on car deceleration, digesting the definition of deceleration and the equations of kinematics is crucial for solving the problem accurately and efficiently.

Additionally, it often helps to visualize the scenario, make sketches, and list given information and required unknowns. Subsequently, as seen in the exercise, equating variables and solving algebraically to find a relation between quantities showcases a systematic approach to physics problems.