How Do You Calculate Distance at Different Speeds Using Integration?

[Blade]

A car traveling at 45 miles per hour is brought to a stop, at constant deceleration, 132 feet from where the brakes are applied.

a. How far has the car moved when its speed has been reduced to 30 miles per hour?

b. How far has the car moved when its speed has been reduced to 15 miles per hour?

My problem here is, I am suppose to use integration (not kinematics). I'm having problems seeing where to start using integration.

 

[AD]

Well, you need to determine the acceleration before you can start. Use v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity and s is the displacement.

Once you've done that, you can express the velocity as a function of time

v = u + at

Integrate to find displacement

s = ut + ½at^2

 

[M98Ranger]

To solve this problem using integration, we first need to understand the concept of acceleration and how it relates to velocity and position. Acceleration is the rate of change of velocity, and it is represented by the derivative of velocity with respect to time. In other words, if we integrate acceleration with respect to time, we can find the change in velocity over a given time period.

In this problem, we are given the initial velocity of the car (45 miles per hour) and the distance it travels until it comes to a complete stop (132 feet). We are also told that the car is decelerating at a constant rate. This means that the acceleration is constant, and we can represent it as a negative value since it is causing the car to slow down.

a. To find how far the car has moved when its speed is reduced to 30 miles per hour, we need to find the change in velocity from 45 miles per hour to 30 miles per hour. This can be represented as ∆v = 30 - 45 = -15 miles per hour. We also know that the car decelerates at a constant rate, so we can represent this change in velocity as the integral of acceleration with respect to time.

∆v = ∫ a dt

Since the car is traveling at constant deceleration, we can substitute the value of acceleration into the integral. Let's say the acceleration is represented by the variable "k" (which will be a negative value).

∆v = ∫ k dt

We can now solve the integral by taking the antiderivative of k with respect to t.

∆v = kt + C

Where C is a constant of integration. Since we are given the initial velocity of the car (45 miles per hour), we can substitute this value into the equation.

45 miles per hour = kt + C

We also know that the initial velocity of the car is 45 miles per hour when t = 0. This means that C = 45.

45 miles per hour = kt + 45

Now, we can solve for k by substituting the final velocity (30 miles per hour) and solving for t.

30 miles per hour = kt + 45

t = -15/k

Substituting this value of t back into the equation, we can solve for k.

30 miles per hour = k(-15/k) + 45