### Video Transcript

A car crashes into a large tree that does not move. The car decelerates from a velocity of 30 meters per second to rest over a distance of 1.3 meters. A driver of mass 70 kilograms is sitting in the car when it crashes. What impulse is applied to the driver by the seat belt assuming he follows the same motion as the car? What is the average force applied to the driver by the seatbelt?

Letβs begin by highlighting some of the important information weβve been given. Weβre told that before it runs into the tree, the car is moving at a speed of 30 meters per second and that it comes to a stop over a distance of 1.3 meters. We want to solve for the impulse applied to the driver by the seat belt; weβll call that impulse capital π½.

That impulse is a vector, having both magnitude and direction. In part two, we wanna solve for the average force applied to the driver by the seat belt; weβll call that capital πΉ. That value is also a vector.

Letβs begin by drawing a picture of the scenario. In our sketch, we show the car approaching the tree at the speed of the π£ sub π, which weβre given as 30 meters per second. Weβve defined motion in the carβs velocity direction as motion in the positive π₯ direction. After the car runs into the tree, it comes to a complete stop over a distance of π equals 1.3 meters. And the mass of the driver in the car what, weβve called π sub π, is equal to 70 kilograms.

Letβs begin part one, solving for impulse, by recalling both the definition of impulse and then the impulse momentum theorem. The impulse applied to an object is defined as the force acting on an object over the time, Ξπ‘, for which that force acts. In our case, for example, the seatbelt exerts an impulse on the driver; that is, is a force over some amount of time.

Impulse is connected with momentum through a theorem called the impulse momentum theorem. This theorem is a long way of saying that the impulse on an object; that, is the force acting on it over the time that force acts is equal to the change in momentum of the object. So for our scenario, we can connect the definition for impulse with this theorem to write that impulse π½ equals πΉΞπ‘, which equals the change in momentum, π, of the object, in this case the driver.

And we can go one step further by recalling the definition of momentum, π. An objectβπ momentum is defined as its mass times its velocity. So the change in momentum, Ξπ, in a nonrelativistic case such as ours is equal to π the mass times of the change in velocity π£. Based on the information given to us, we know π and we can solve for Ξπ£.

What weβve written as π is in our case π sub π, the mass of the driver. And Ξπ£ is equal to the final velocity on the car, π£ sub π, minus the initial velocity. Since the car ultimately comes to a rest, π£ sub f is zero and π£ sub i is equal to 30 meters per second in the positive π₯ direction.

We can write that as 30π meters per second. Now that we know Ξπ£, weβre ready to solve for the impulse delivered to the driver by the seat belt, π½. π½ equals π sub π times Ξπ£ or 70 kilograms multiplied by negative 30π meters per second. Multiplying these numbers together, to two significant figures, the impulse acting on the driver is negative 2.1 times 10 to the third π kilogram-meters per second. Thatβs the impulse the seat belt delivers to the driver.

Now that weβve solved for the impulse delivered to the driver, what about the average force that the driver experiences from the seat belt? Looking at our impulse definition, we can use this equation to solve for that force by dividing through both sides by Ξπ‘. This gives us an equation which says that the average force acting on the driver, πΉ, equals the impulse acting on the driver, π½, divided by the time over which the force was acting.

We know π½, the impulse, from having solved for it. But what about Ξπ‘? Letβs consider again how our car came to a stop. The car had an initial speed, 30 meters per second, and a final speed of zero meters per second, and this stop happened over a distance of 1.3 meters. Weβre solving for average force πΉ, and if we assume that the acceleration of the car over its stop was constant, that means we can use the kinematic equations to solve for Ξπ‘.

Letβs recall those equations. Of these four equations, weβre looking for one that matches the information weβve been given as well as the information weβre searching for, time. The fourth equation is a good match. Applying it to our scenario: π is a given value, 1.3 meters; π£ zero weβve called π£ sub π, and weβre given that as well; π£ sub π, the final speed of the car, is zero; and π‘ weβll call Ξπ‘, the time over which this change in speed happens.

In a simplified form, π is equal to the initial speed π£ sub π divided by two times Ξπ‘. Multiplying both sides by two over π£ sub π, those terms cancel out of the right-hand side of our equation. And we see the Ξπ‘ is equal to two times π over π£ sub π. When we plug in for π and π£ sub π and calculate this fraction, we find a Ξπ‘ values of 0.087 seconds.

So with that being Ξπ‘, weβre now ready to solve for the force acting on the driver from the seat belt, πΉ. For the impulse π½, weβll substitute in negative 2.1 times 10 to the third π kilogram-meters per second; and for Ξπ‘, weβll substitute 0.087 seconds. When we calculate this fraction, to two significant figures, we find a force value of negative 24 times 10 to the third π newtons. This is the average force the seat belt exerts on the driver.