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Chapter 2: Problem 54

A car accelerates at \(2.0 \mathrm{m} / \mathrm{s}^{2}\) along a straight road. It passes two marks that are \(30 \mathrm{m}\) apart at times \(t=4.0 \mathrm{s}\) and \(t=5.0 \mathrm{s}\) What was the car's velocity at \(t=0\) s?

Short Answer

Expert verified
Thus, the car's velocity at \(t=0\) was \(-6m/s\).

Step by step solution

01

Determine the final velocity

Given that the car passes two marks that are 30m apart at times \(t=4.0s\) and \(t=5.0s\), this implies that the time taken to cover 30m is \((5.0s - 4.0s) = 1.0s\). Since the car was accelerating at \(2.0 m/s^{2}\), it implies that the change in velocity over this time period, which is also the final velocity is \(acceleration \times time = (2.0 m/s^{2}) \times 1.0s = 2m/s\). Thus, the final velocity of the car is \(2m/s\).
02

Calculate the initial velocity

Next, use the third equation of motion to solve for the initial velocity. Rearranging the equation, we get \(u = v - at\). Substituting the given values we get \(u = 2m/s - (2m/s^{2} \times 4s) = 2m/s - 8m/s = -6m/s\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equations of Motion
When working with moving objects, the equations of motion provide a vital set of tools for predicting an object鈥檚 future position, velocity, and acceleration at any given time. These equations link an object's initial velocity, final velocity, acceleration, time, and displacement. One such equation is the third equation of motion, which is often expressed as
\( v^2 = u^2 + 2as \)
where
  • \( v \) is the final velocity,
  • \( u \) is the initial velocity,
  • \( a \) is the acceleration, and
  • \( s \) is the displacement.
However, in the provided exercise, the equation has been rearranged to solve for the initial velocity
(\( u \)), resulting in the formula
\( u = v - at \).

Understanding the Third Equation of Motion

In situations where constant acceleration is involved, you can use the third equation of motion to find any of the variables if the other three are known. This equation is particularly useful for kinematics problems where time is not a factor. By manipulating the equation, you can either solve for the initial velocity (as shown in the exercise), final velocity, time, displacement, or acceleration, making it a versatile tool for analyzing motion. Solving kinematics problems often involves selecting the appropriate equation of motion and rearranging it to find the missing variable.
Acceleration
Acceleration is defined as the rate at which the velocity of an object changes with time. It is a vector quantity, meaning it has both a magnitude and a direction. In SI units, it is measured in meters per second squared (\(m/s^2\)). The concept of acceleration is central to many physics problems, including kinematics, where we are concerned with the motion of objects.

Constant Acceleration in Kinematics

In many problems, we assume that acceleration is constant. This assumption simplifies calculations and is a good approximation for many real-world scenarios, such as the acceleration of a car on a straight road. In the exercise example, the car鈥檚 acceleration is given as \(2.0 m/s^2\), indicating that for every second the car is moving, its velocity increases by \(2.0 m/s\). Understanding acceleration is key to solving kinematics problems, as it allows us to connect an object's initial velocity, final velocity, and displacement.
Initial Velocity
The term 'initial velocity' refers to the velocity of an object before it has experienced acceleration. It's an essential component in kinematics equations and can be thought of as the velocity at the starting point of your observation. In kinematics, initial velocity provides a basis for predicting future motion when combined with other factors like time and acceleration.

Determining Initial Velocity

To determine the initial velocity in kinematics, we often use one of the kinematic equations, depending on the given variables. For the given exercise, initial velocity is found using a rearranged version of the third equation of motion, and this step is crucial in outlining the motion trajectory of the vehicle. Initial velocity doesn鈥檛 just indicate a starting speed鈥攊t serves as the foundation from which changes in velocity are gauged, due to acceleration or deceleration within a specific time frame.

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Most popular questions from this chapter

Heather and Jerry are standing on a bridge 50 m above a river. Heather throws a rock straight down with a speed of \(20 \mathrm{m} / \mathrm{s}\) Jerry, at exactly the same instant of time, throws a rock straight up with the same speed. Ignore air resistance. a. How much time elapses between the first splash and the second splash? b. Which rock has the faster speed as it hits the water?

You are driving to the grocery store at \(20 \mathrm{m} / \mathrm{s}\). You are \(110 \mathrm{m}\) from an intersection when the traffic light turns red. Assume that your reaction time is \(0.50 \mathrm{s}\) and that your car brakes with constant acceleration. a. How far are you from the intersection when you begin to apply the brakes? b. What acceleration will bring you to rest right at the intersection? c. How long does it take you to stop after the light turns red?

A driver has a reaction time of \(0.50 \mathrm{s}\), and the maximum deceleration of her car is \(6.0 \mathrm{m} / \mathrm{s}^{2} .\) She is driving at \(20 \mathrm{m} / \mathrm{s}\) when suddenly she sees an obstacle in the road \(50 \mathrm{m}\) in front of her. Can she stop the car in time to avoid a collision?

A cat is sleeping on the floor in the middle of a 3.0 -m-wide room when a barking dog enters with a speed of \(1.50 \mathrm{m} / \mathrm{s}\). As the dog enters, the cat (as only cats can do) immediately accelerates at \(0.85 \mathrm{m} / \mathrm{s}^{2}\) toward an open window on the opposite side of the room. The dog (all bark and no bite) is a bit startled by the cat and begins to slow down at \(0.10 \mathrm{m} / \mathrm{s}^{2}\) as soon as it enters the room. Does the dog catch the cat before the cat is able to leap through the window?

A \(1000 \mathrm{kg}\) weather rocket is launched straight up. The rocket motor provides a constant acceleration for \(16 \mathrm{s}\), then the motor stops. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance. a. What was the rocket's acceleration during the first 16 s? b. What is the rocket's speed as it passes through a cloud \(5100 \mathrm{m}\) above the ground?
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