Question

A car is traveling due west at \(20.0 \mathrm{~m} / \mathrm{s}\). Find the velocity of the car after \(3.00 \mathrm{~s}\) if its acceleration is \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) due west. 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

Answer: The final velocity of the car is \(23.0 \mathrm{~m} / \mathrm{s}\) west.

Step-by-step solution

List the given information

We are given: Initial velocity (v0) = \(20.0 \mathrm{~m} / \mathrm{s}\) (due west) Acceleration (a) = \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) (due west) Time (t) = \(3.00 \mathrm{~s}\)

Apply the formula for velocity under constant acceleration

The formula is: \(v = v_0 + at\), where: v = final velocity v0 = initial velocity a = acceleration t = time

Plug in the given values and calculate the final velocity

Now, we'll substitute the given values into the formula: \(v = 20.0 \mathrm{~m} / \mathrm{s} \ + (1.0 \mathrm{~m} / \mathrm{s}^{2})(3.00 \mathrm{~s})\) \(v = 20.0 \mathrm{~m} / \mathrm{s} \ + (1.0 \mathrm{~m} / \mathrm{s}^{2})(3.00 \mathrm{~s}) = 20.0 \mathrm{~m} / \mathrm{s} + 3.0 \mathrm{~m} / \mathrm{s} = 23.0 \mathrm{~m} / \mathrm{s}\) The final velocity of the car is \(23.0 \mathrm{~m} / \mathrm{s}\) due west. The correct answer is: c) \(23.0 \mathrm{~m} / \mathrm{s}\) west

Key concepts

Velocity

Velocity is a fundamental concept in kinematics. It describes the speed of an object in a specific direction. Unlike speed, which is scalar, velocity is a vector quantity. This means it includes both magnitude (how fast something is moving) and direction (where it is heading). When you say a car is moving at 20 meters per second west, that's a velocity.Understanding velocity is crucial because it allows us to predict how the position of an object changes over time. We can use it to calculate how far an object will travel in a given direction.

• Initial velocity: the velocity at which an object starts, denoted as \(v_0\).
• Final velocity: the velocity of an object at the end of a time interval, denoted as \(v\).

Knowing both the initial and final velocities along with the time and acceleration helps in solving many physics problems.

Acceleration

Acceleration is the rate at which an object's velocity changes over time. If an object is speeding up, slowing down, or changing direction, it is accelerating. Like velocity, acceleration is a vector quantity and has both magnitude and direction.The formula for acceleration is \(a = \frac{\Delta v}{t}\), where \(\Delta v\) is the change in velocity and \(t\) is the time it takes for this change to occur. In the context of the exercise, the car accelerates at \(1.0 \, \mathrm{m/s^2}\) west:

• A positive acceleration in the direction of initial velocity increases speed.
• A negative acceleration (deceleration) or acceleration opposite to the direction decreases speed.

Understanding acceleration is essential for analyzing and predicting how quickly an object’s speed and direction can change.

Constant Acceleration

Constant acceleration occurs when an object's acceleration remains the same over time. This simplification is essential in physics because it allows us to use straightforward mathematical equations to predict motion.When acceleration is constant, the following kinematic equation can be used: \(v = v_0 + at\). In this situation:

• \(v\) is the final velocity.
• \(v_0\) is the initial velocity.
• \(a\) is the constant acceleration.
• \(t\) is the time.

For example, if a car speeds up with a constant acceleration of \(1.0 \, \mathrm{m/s^2}\) west, after 3 seconds the velocity increase can be directly calculated by multiplying the acceleration by the time and adding it to the initial velocity. Constant acceleration simplifies complex motion analysis and is a vital part of kinematics, especially when trying to predict an object's future motion.