# Video: Finding the Force Required for Deceleration of an Object over a Particular Distance

A car with a mass of 1000 kg applies brakes when it is moving at 90.0 km/h, and the car comes to a rest after travelling 40.0 m. What is the net force on the car during its deceleration?

06:34

### Video Transcript

A car with a mass of 1000 kilograms applies brakes when it is moving at 90.0 kilometers per hour, and the car comes to a rest after travelling 40.0 meters. What is the net force on the car during its deceleration?

To start off, let’s draw a diagram that shows the car at its initial speed and then when it’s come to a stop. So here we have our car starting out with an initial speed that we’ll call 𝑣 sub 𝑖 of 90 kilometers per hour. And then over a distance of 40.0 meters, the car ends up with the final speed, we’ll call it 𝑣 sub 𝑓, of zero. It’s come to a stop. And we want to figure out what is the force, we’ll call it capital 𝐹, that is applied to the car over this breaking process so that it comes to a stop after having started with this initial speed.

Now you may guess that this is a Newton’s second law problem. And if so, you’re exactly right. So let’s get that set up to figure out what the force on this car is. Alright. Now, recall that Newton’s second law states this: The net force acting on an object is equal to the mass of that object multiplied by its acceleration. So in the case of our problem, our first order of business is to start working with the given speeds of this car at the beginning and end of its journey. In particular, we’d like to take this initial speed that we’re given and convert the units of that speed to something that will let us work with meters per second rather than kilometers per hour. So let’s work on that now.

Okay. So the initial speed of our car is 90.0 kilometers per hour. But, to get this into units that work well with newtons — and remember that the units of a newton, the unit of force, are kilogram meters per second squared — so to get our initial velocity in those terms, let’s multiply it through to remove kilometers per hour and replace it with meters per second. Let’s start by converting the hours in the denominator to seconds. To do that, we can multiply by the fraction which says that one hour is equal to 3600 seconds. Looking at that hour’s unit, we recall that one hour is equal to 3600 seconds. And going on from there, we know that one kilometer is equal to 1000 meters.

So looking now at our expression for the initial speed, let’s cancel out the units we can. We see the h representing hours cancel out, and we also see the kilometers cancel out leaving us with meters per second, the units we were seeking. When we multiply all these numbers through, we find that we have an initial velocity in terms of meters per second of 25.0 meters per second. Great. Now that we have that number and the units we prefer, we can start to look through the kinematic equations to figure out what is the acceleration of this car over the course of its stop. So let’s turn to look at those kinematic equations now. As we look through these different equations, we see that in the second kinematic equation, we have an expression that relates our final velocity, which we know in our case, to the initial velocity plus the acceleration, which we’re seeking for, multiplied by a distance. And again, we’re given the distance in this problem. So this equation looks like it might fit our situation very well.

Okay. Let’s take a look at how this equation applies to our given variables. So our final speed of our car, we know is zero. So this initial term here is zero. Our initial speed of the car, we just calculated out in meters per second, is 25.0; so that’s known. Then we have acceleration that we’re looking to solve for. And 𝑑, the distance the car takes to stop, is also known, given as 40.0 meters. So we have everything we need to rearrange this equation and solve for the acceleration of the car as it comes to a stop. Rewriting this equation in light of the fact that our car ends at a rest, we can subtract 𝑣 sub 𝑖 squared from both sides. And that will cancel out the 𝑣 sub 𝑖 squared on the right side of the equation. Then we can go ahead and divide both sides of the equation by two times 𝑑, the distance the car takes to come to a stop. This will cancel those terms out on the right side leaving us with 𝑎, the acceleration we’re seeking.

Let’s plug in numbers for 𝑣 sub 𝑖 and 𝑑 to solve numerically for 𝑎. Negative 25.0 meters per second quantity squared divided by two times 40.0 meters is equal to negative 7.81 meters per second squared. That’s the acceleration our car experiences as it comes to a stop.

Now let’s go back to Newton’s second law to solve ultimately for the force that the car experiences. So we now know the acceleration the car experienced. We were given its mass of 1000 kilograms and we recall Newton’s second law which says that the force the car experiences is equal to the product of these two variables. Multiplying 1000 kilograms for mass times negative 7.81 meters per second squared for acceleration, we find that the force on the car equals 7810 newtons, which is the same as negative 7.81 kilo newtons. That is the net force the car experiences as it comes to a stop.