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A car is traveling due west at \(20.0 \mathrm{~m} / \mathrm{s}\). Find the velocity of the car after \(37.00 \mathrm{~s}\) if its constant acceleration is \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) due east. Assume the acceleration remains constant. a) \(17.0 \mathrm{~m} / \mathrm{s}\) west b) \(17.0 \mathrm{~m} / \mathrm{s}\) east c) \(23.0 \mathrm{~m} / \mathrm{s}\) west d) \(23.0 \mathrm{~m} / \mathrm{s}\) east e) \(11.0 \mathrm{~m} / \mathrm{s}\) south

Short Answer

Expert verified
Answer: b) 17.0 m/s east

Step by step solution

01

Calculate the cumulated velocity due to acceleration

To compute the accumulated velocity due to the acceleration, use the formula: \(V_{acc} = a \cdot t\) where \(V_{acc}\) is the accumulated velocity, \(a\) is the acceleration, and \(t\) is the time duration. Given that, \(a = 1.0 \mathrm{~m} / \mathrm{s}^{2}\) and \(t = 37.00 \mathrm{~s}\), \(V_{acc} = 1.0 \mathrm{~m} / \mathrm{s}^{2} \cdot 37.00 \mathrm{~s} = 37.0 \mathrm{~m} / \mathrm{s}\)
02

Determine the final velocity

We know that the initial velocity (\(V_{initial}\)) is \(20.0 \mathrm{~m} / \mathrm{s}\) west, and the accumulated velocity due to acceleration is \(37.0 \mathrm{~m} / \mathrm{s}\). Since both the initial velocity and the accumulated velocity due to acceleration are in opposite directions (west and east), we subtract the accumulated velocity from the initial velocity to find the final velocity: \(V_{final} = V_{initial} - V_{acc}\) \(V_{final} = 20.0 \mathrm{~m} / \mathrm{s} - 37.0 \mathrm{~m} / \mathrm{s}\) \(V_{final} = -17.0 \mathrm{~m} / \mathrm{s}\) Since the result is negative, it means the final velocity is in the opposite direction to the initial velocity, which is east. So, the final velocity of the car after \(37.00 \mathrm{~s}\) is \(17.0 \mathrm{~m} / \mathrm{s}\) east. The correct answer is option b) \(17.0 \mathrm{~m} / \mathrm{s}\) east.

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause this motion. It helps us understand and predict how objects move under different circumstances. Kinematic equations are essential tools used to describe the motion of objects along a straight line, usually expressed in terms of velocity, acceleration, and time. In the context of the problem with the car traveling west, kinematics helps us predict what happens when a constant acceleration is applied in the opposite direction.

The key kinematic equations allow you to connect different variables like initial velocity, final velocity, acceleration, and time. They make it easier to solve real-world problems involving motion. In summary, kinematics lays the groundwork for understanding any motion-related problems by offering a set of equations and concepts that model how objects move.
Velocity Calculation
Velocity is a vector quantity, which means it has both magnitude and direction. Unlike speed, which only measures how fast something is moving, velocity tells us how fast and in what direction an object is moving. Calculating velocity is pivotal for understanding motion, as it allows us to evaluate how an object’s position changes with time.

In the given exercise, we begin with an initial velocity of the car, given as \(20.0 \mathrm{~m/s}\) in the west direction. The problem statement asks us to determine the final velocity after a constant acceleration is applied. This involves calculating the change in velocity due to acceleration and then adjusting the initial velocity.

The formula used is:
\(V_{final} = V_{initial} + V_{acc}\)
but since directions differ, we effectively subtract:
\(V_{final} = V_{initial} - V_{acc}\)

By calculating properly, we find out not only the numeric value of the velocity but also its new direction, showing the total influence of the applied acceleration.
Constant Acceleration
Acceleration is defined as the rate of change of velocity over time, and when this rate is constant, it simplifies many calculations in physics. Constant acceleration means that an object's velocity changes by the same amount each second. This concept is highly significant because it makes predictions about how an object moves over time straightforward.

In problems like the one with the car, where acceleration is constant, we can use established kinematic formulas like:
\(V_{acc} = a \times t\)
Using this formula, we calculate how much the velocity changes. In our example, the car has a constant acceleration of \(1.0 \mathrm{~m/s^2}\) east, affecting its velocity over a period of \(37.00 \mathrm{~s}\).

The simplicity of constant acceleration allows us to directly relate time, acceleration, and change in velocity using a simple multiplication, making prediction of future states feasible and quite manageable in physics.

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Most popular questions from this chapter

The Bellagio Hotel in Las Vegas, Nevada, is well known for its Musical Fountains, which use 192 HyperShooters to fire water hundreds of feet into the air to the rhythm of music. One of the HyperShooters fires water straight upward to a height of \(240 \mathrm{ft}\). a) What is the initial speed of the water? b) What is the speed of the water when it is at half this height on its way down? c) How long will it take for the water to fall back to its original height from half its maximum height?

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A speeding motorcyclist is traveling at a constant speed of \(36.0 \mathrm{~m} / \mathrm{s}\) when he passes a police car parked on the side of the road. The radar, positioned in the police car's rear window, measures the speed of the motorcycle. At the instant the motorcycle passes the police car, the police officer starts to chase the motorcyclist with a constant acceleration of \(4.0 \mathrm{~m} / \mathrm{s}^{2}\) a) How long will it take the police officer to catch the motorcyclist? b) What is the speed of the police car when it catches up to the motorcycle? c) How far will the police car be from its original position?

A girl is standing at the edge of a cliff \(100 . \mathrm{m}\) above the ground. She reaches out over the edge of the cliff and throws a rock straight upward with a speed \(8.00 \mathrm{~m} / \mathrm{s}\). a) How long does it take the rock to hit the ground? b) What is the speed of the rock the instant before it hits the ground?

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