Question

A car traveling at \(25.0 \mathrm{~m} / \mathrm{s}\) applies the brakes and decelerates uniformly at a rate of \(1.2 \mathrm{~m} / \mathrm{s}^{2}\) a) How far does it travel in \(3.0 \mathrm{~s}\) ? b) What is its velocity at the end of this time interval? c) How long does it take for the car to come to a stop? d) What distance does the car travel before coming to a stop?

Short answer

Question: A car is decelerating uniformly from an initial velocity of 25.0 m/s. The uniform deceleration is 1.2 m/s². Calculate (a) the distance traveled in 3.0 s, (b) its velocity at the end of the time interval, (c) the time taken for the car to come to a stop, and (d) the distance traveled before coming to a stop. Answer: (a) The distance traveled in 3.0 s is 63.0 m. (b) The car's velocity at the end of the 3.0 s time interval is 21.4 m/s. (c) It takes 20.8 s for the car to come to a stop. (d) The car travels a distance of 260.0 m before coming to a stop.

Step-by-step solution

Part (a): Distance traveled in 3.0 s

To find the distance traveled in 3.0 s, we will use the displacement equation: \(s = ut + \frac{1}{2}at^2\) Plugging in the given values: \(s = (25.0)(3.0) + \frac{1}{2}(-1.2)(3.0)^2\) Solve for s: \(s = 63.0 \mathrm{~m}\) The car travels a distance of 63.0 m in 3.0 s.

Part (b): Velocity at the end of the time interval

Now we will find the final velocity after 3.0 s using the second equation of motion: \(v = u + at\) Plugging in the given values: \(v = 25.0 + (-1.2)(3.0)\) Solve for v: \(v = 21.4 \mathrm{~m/s}\) The car's velocity at the end of the time interval is 21.4 m/s.

Part (c): Time taken for the car to come to a stop

To find the time it takes for the car to come to a stop, we'll use the second equation of motion again: \(v = u + at\) In this case, the final velocity (v) should be zero as the car comes to a stop. Therefore: \(0 = 25.0 - 1.2t\) Solve for t: \(t = 20.8 \mathrm{~s}\) It takes 20.8 s for the car to come to a stop.

Part (d): Distance traveled before coming to a stop

Now that we know the time it takes for the car to stop, we can find the distance using the first equation of motion: \(s = ut + \frac{1}{2}at^2\) Plugging in the known values: \(s = (25.0)(20.8) + \frac{1}{2}(-1.2)(20.8)^2\) Solve for s: \(s = 260.0 \mathrm{~m}\) The car travels a distance of 260.0 m before coming to a stop.

Key concepts

Uniform Deceleration

When we say a vehicle experiences uniform deceleration, we are referring to a constant rate at which the vehicle's speed decreases over time. This consistent rate is crucial for solving kinematic problems because it allows us to apply the laws of motion in a straightforward way. In our exercise, the car slows down at a uniform deceleration rate of \(1.2 \mathrm{~m} / \mathrm{s}^{2}\), meaning that every second, its velocity reduces by \(1.2 \mathrm{~m/s}\).

Understanding this uniform change in speed helps students in predicting and calculating various aspects of the car's journey, such as how long it will take to come to a complete stop or how far it will travel during deceleration. It's akin to applying the brake pedal with unwavering pressure in a real-world scenario. Uniform deceleration contrasts with non-uniform deceleration, where the rate of slowing down changes over time, making computations more complex.

Equations of Motion

When we analyze the motion of an object, such as our decelerating car, we rely on the equations of motion to describe its behavior mathematically. These equations enable us to connect the dots between the car's initial velocity, acceleration (or deceleration in this case), and the time involved to determine the final velocity and displacement.

For example, in our exercise, the second equation of motion, \(v = u + at\), helped us find the final velocity after a certain time with a known deceleration value. Similarly, the displacement equation, \(s = ut + \frac{1}{2}at^2\), gave us the car's displacement over that time period. These formulas are pivotal elements of kinematics, helping students to derive meaningful conclusions from basic data points about an object's movement. It is important to recognize which formula to apply in different scenarios to find the correct solution efficiently.

Distance and Displacement

Distance and displacement may seem similar, but in the realm of physics, they serve distinct purposes. Distance is a scalar quantity that refers to the total path length covered by the car without concern for its direction. Displacement, on the other hand, is a vector quantity, not only considering the length traveled but also in which direction the car has moved.

In kinematic problems like the one we've tackled, these concepts help us determine the covered path and the car's change in position. It's important to know when to calculate for distance and when displacement is required. In part (a) of our exercise, we calculated the distance traveled by the car in a specific timeframe using the displacement formula because the car was moving in a straight line, so here the distance and displacement are equal. When solving these types of problems, picturing the scenario can aid in better grasping which quantity to compute for a correct understanding of the car's motion.