Question

The driver of a car traveling at \(25.0 \mathrm{~m} / \mathrm{s}\) applies the brakes, and the car decelerates uniformly at a rate of \(1.2 \mathrm{~m} / \mathrm{s}^{2}\). a) How far does it travel in 3.0 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 traveling at an initial velocity of 25.0 m/s and decelerates at a rate of 1.2 m/s². Calculate the following: a) the distance traveled in 3.0 seconds, b) the velocity at the end of 3.0 seconds, c) the time required for the car to come to a stop, and d) the total distance covered before coming to a stop. Answer: a) The car travels 26.4 meters in 3.0 seconds. b) The car's velocity at the end of 3.0 seconds is 21.4 m/s. c) It takes 20.83 seconds for the car to come to a stop. d) The car covers 260.42 meters before coming to a stop.

Step-by-step solution

Identify the known variables

We know the initial velocity (u) is 25.0 m/s, the deceleration (a) is -1.2 m/s^2 (negative because it's slowing down), and the time interval (t) is 3.0 seconds.

Use equation 2 to calculate the distance traveled

We will use the formula \(s = ut + \frac{1}{2}at^2\). Plug in the known values: \(s = (25.0)(3.0) + \frac{1}{2}(-1.2)(3.0)^2\).

Calculate the distance

Simplify the expression: \(s = 75.0 - 5.4(9) = 75.0 - 48.6 \Rightarrow s = 26.4 \mathrm{~m}\). The car travels 26.4 meters in 3.0 seconds. #b) Velocity at the end of 3.0 seconds#

Use equation 1 to calculate the final velocity

We will use the formula \(v = u + at\). Plug in the known values: \(v = 25.0 + (-1.2)(3.0)\).

Calculate the velocity

Simplify the expression: \(v = 25.0 - 3.6 \Rightarrow v = 21.4 \mathrm{~m} / \mathrm{s}\). The car's velocity after 3.0 seconds is 21.4 m/s. #c) Time for the car to come to a stop#

Use equation 1 with final velocity as 0

Since the car will come to a stop, the final velocity (v) will be 0 m/s. So, use equation 1 as: \(0 = 25.0 + (-1.2)t\).

Calculate the time

Solve for t: \(\frac{-25.0}{-1.2} = t \Rightarrow t = 20.83 \mathrm{~s}\). It takes 20.83 seconds for the car to come to a stop. #d) Distance traveled before coming to a stop#

Use equation 3 with final velocity as 0

Since the car will come to a stop, the final velocity (v) will be 0 m/s. So, use equation 3 as: \(0^2 = (25.0)^2 + 2(-1.2)s\).

Calculate the distance

Solve for s: \(s = \frac{625}{2.4} \Rightarrow s = 260.42 \mathrm{~m}\). The car covers 260.42 meters before coming to a stop.

Key concepts

Uniform Deceleration

Uniform deceleration occurs when an object slows down at a constant rate. In our exercise, a car traveling at 25.0 m/s decelerates uniformly, which means its deceleration rate is constant at -1.2 m/s². This negative value indicates a decrease in speed. One key application of this concept is in safety calculations, such as determining the stopping distance for vehicles, ensuring they can come to a safe halt within a given distance.

It's important to note that 'uniform' implies a steady reduction in speed, and this consistency simplifies our calculations. When we're told that the car decelerates uniformly, we can use kinematic equations confidently, knowing that the acceleration rate doesn't change over time. This helps us predict the car's behavior over specific intervals.

Distance-Time Relationship

The distance-time relationship in kinematics allows us to compute the displacement of an object over a period. The equation used in the exercise, s = ut + (1/2)at², combines initial velocity (u), acceleration (a), and time (t) to calculate the distance (s).

In this problem, the distance the car travels in 3.0 seconds under uniform deceleration is determined by plugging the values into the formula. This calculation shows the clear relationship between time and distance when acceleration is factored in. The remarkable aspect of this formula is its ability to accurately chart an object's journey, even when slowing down, as each moment passes.

Final Velocity Calculation

Final velocity is a critical component in kinematics, representing the speed of an object at the end of a time interval. To calculate the final velocity in the case of our car, we use the equation v = u + at. This equation factors in the initial velocity (u), acceleration (a), and time (t) to give us the final velocity (v).

Even with deceleration, the final velocity after a certain time can be found by substituting the initial velocity and the product of deceleration with time into this equation. It’s valuable in scenarios such as traffic management, where knowing a vehicle's speed after a certain response time can inform us about the effectiveness of braking systems and the overall safety of the vehicle.

Stopping Time and Distance

Stopping time refers to the duration it takes for a moving object to come to a complete stop under a uniform deceleration. Similarly, stopping distance is the total distance an object covers from the moment deceleration begins to when it stops completely. The stopping time is extracted using the initial velocity and deceleration, whereas the stopping distance is calculated using kinematic equations tailored to situations where the final velocity is zero.

These measures are essential in designing safe automotive and transportation systems. For instance, determining the minimum distance needed for a vehicle to stop safely is a direct application of both stopping time and distance concepts, and is critical for road safety regulations.