You are arguing over a cell phone while trailing an unmarked...

You are arguing over a cell phone while trailing an unmarked police car by $25 \mathrm{~m} ;$ both your car and the police car are traveling at $110 \mathrm{~km} / \mathrm{h}$. Your argument diverts your attention from the police car for $2.0 \mathrm{~s}$ (long enough for you to look at the phone and yell, " 1 won't do that $\left.\right|^{n}$ ). At the beginning of that $2.0 \mathrm{~s}$. the police officer begins braking suddenly at $5.0 \mathrm{~m} / \mathrm{s}^{2}$. (a) What is the separation between the two cars when your attention finally returns? Suppose that you take another $0.40 \mathrm{~s}$ to realize your danger and begin braking. (b) If you too brake at $5.0 \mathrm{~m} / \mathrm{s}^{2}$, what is your speed when you hit the police car?

Key Concepts

Uniform Acceleration

Uniform acceleration refers to motion in which an object's velocity changes at a constant rate. This concept is essential for analyzing situations where vehicles are braking, as both the police car and the pursuing car decelerate at constant rates, allowing the use of standard equations to predict their future positions and speeds.

Reaction Time

Reaction time is the interval between the onset of a stimulus and the initiation of a response. In driving scenarios, delays in reaction time contribute to additional distance traveled before braking commences, making it a critical factor in safety calculations and collision avoidance analyses.

Relative Motion

Relative motion involves assessing the movement of one object from the viewpoint of another moving object. In problems that determine separation between vehicles or the outcome of an impending collision, considering the relative velocities and positions is key to accurately predicting when and where one object will meet or impact another.

Kinematic Equations

Kinematic equations provide mathematical relationships between displacement, velocity, acceleration, and time for objects moving with constant acceleration. These equations are used to calculate distances traveled during reaction times and deceleration phases, making them fundamental tools for solving vehicle-braking and collision problems.

Transcript

00:01 In this problem of motion in a straight line, we have given that you are arguing over a cell phone while trailing an unmarked police car by 25 meters.

00:12 So this is 25 meters.

00:16 Both your car and the police car are traveling at 110 kilometers per hour.

00:23 Your arguments diverts your attention from the police car for 2 seconds.

00:29 So, this is 2 .0 seconds.

00:34 Long enough for you to look at the phone and yell.

00:39 I don't do that.

00:42 At the beginning of 2 seconds, the police officer begins breaking suddenly at 5 meters per second.

00:50 So, this is 5 meters per second square.

00:53 This is the acceleration or we can say this is breaking.

00:55 So this will be deceleration.

00:58 Now we have to find what is the separation between the two cars when your attention? finally returns and also we have to suppose that you take another 0 .40 seconds to realize your danger and being and begin breaking and in the part b if you two break at same acceleration or we can say if we are waking so this would be desaleration 5 .0 meters per second square what is your speed change or we can say what is your speed when you hit the police car.

01:40 So first this is for part a we have to convert this speed this is 110 kilometers per hour when this converted into meters per second so this is equal to 30 .56 meters per second and during a two second interval say here during two second interval you would travel to multiple 30 .5 6 meter per second that means you travel 61 .11 meters and the desaltering police car travels now say this is x1 now x2 this is a distance traveled in two second by desalating police car so this value would be this is 30 .56 minus half multiplied with the acceleration so this is 5 .0 and multiplied with time this is 2 square now x2 is equal to and here this value is also multiplied with 2.

02:43 So this would be like this minus.

02:49 Now when you solve it this is equal to 51 .11 so this is 51 .11 meters.

02:59 Now in light of the fact that initial gap was 25 meters this means the gap has been narrowed by 10 meters that is to a distance of 15 meters between the cut.

03:15 Now, the separation becomes, say this is police, this is your car and now separation would become 25 minus 10 that is equal to 15 meters.

03:29 So we have the right answer as 15 meters.

03:33 So this is the answer for part a.

03:35 Now for part b, if you two brake at 5 meters per second, what is your speed when you hit the police car? so now we have some reaction time, this is 0 .4.

03:50 So, after 2 .4 seconds we would apply the brake and during 2 .4 seconds we would travel, this is 30 .56 multiplied by 2 .4 which is equal to 73 .33 meters.

04:09 We would travel this distance...

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