PHYSICS - Calculating the Braking Distance of a Car (Exam Question Example)

Summary

In this educational video, the presenter tackles a physics problem involving a car traveling at 90 km/h that suddenly encounters a barrier 40 meters ahead. The driver takes 0.75 seconds to react before applying brakes, decelerating at 10 m/s². The script explores whether the car will hit the barrier and calculates the maximum safe speed to avoid a collision. Through clear visualization and step-by-step calculations, the presenter finds that at the initial speed, the car will not stop in time, and determines the critical speed to be 14.26 m/s for a safe stop, emphasizing the importance of practice in mastering physics.

Takeaways

• 🚗 The scenario involves a car traveling at 90 km/h that suddenly sees a barrier 40 meters ahead.
• ⏱ The driver takes 0.75 seconds to react before applying the brakes.
• 🔢 The car's initial speed is converted to 25 m/s for consistency in units.
• 🛑 The car decelerates at a rate of 10 m/s² once the brakes are applied.
• 📐 The problem is divided into two phases: the reaction time phase and the deceleration phase.
• 🔍 The first phase (reaction time) is calculated by multiplying the initial speed by the reaction time.
• 📉 The second phase (deceleration) uses the kinematic equation to find the stopping distance.
• 🚫 The calculation shows that the car will not be able to stop in time and will hit the barrier.
• 🔄 For part B, the process is reversed to find the maximum safe speed to avoid hitting the barrier.
• 🧩 The maximum speed is found by setting up a quadratic equation based on the stopping distance and solving for the initial velocity.
• 🏁 The maximum safe speed calculated is approximately 14.26 m/s, which is less than the initial speed of 25 m/s.

Q & A

What is the initial speed of the car in meters per second?

The initial speed of the car is 25 meters per second, which is obtained by converting 90 kilometers per hour to meters per second using the conversion factor \( \frac{1}{3.6} \).

How long does it take for the driver to react before applying the brakes?

The driver takes 0.75 seconds to react before applying the brakes after seeing the barrier.

What is the deceleration rate of the car when the brakes are applied?

The car decelerates at a rate of 10 meters per second squared.

How much distance does the car cover during the driver's reaction time?

During the driver's reaction time, the car covers 18.75 meters.

What is the formula used to calculate the distance covered while decelerating?

The formula used to calculate the distance covered while decelerating is \( v^2 = u^2 + 2as \), where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the distance.

Does the car hit the barrier according to the scenario described?

Yes, the car hits the barrier because the total distance covered during the reaction time and while decelerating exceeds the 40 meters to the barrier.

What is the maximum speed at which the car could travel and not hit the barrier, given the driver's reaction time and the deceleration rate?

The maximum speed at which the car could travel and not hit the barrier is 14.26 meters per second.

How is the total stopping distance divided in the scenario?

The total stopping distance is divided into two parts: the distance covered during the driver's reaction time (s1) and the distance covered while decelerating (s2).

What mathematical approach is used to solve for the maximum speed in Part B of the question?

In Part B, a quadratic equation is set up and solved to find the maximum initial speed (x) that allows the car to stop just before the barrier.

Why is it important to visualize the scenario when solving the physics problem?

Visualizing the scenario helps to break down the problem into manageable parts and understand the sequence of events, making it easier to apply the correct physics formulas and solve the problem accurately.

What is the significance of converting all units to standard units before solving the problem?

Converting all units to standard units, such as meters per second for speed, ensures consistency in calculations and avoids errors due to unit mismatches.

Outlines

00:00

🚗 Car Stopping Distance Calculation

The paragraph discusses a physics problem involving a car traveling at 90 km/h that suddenly encounters a barrier 40 meters ahead. The driver takes 0.75 seconds to react and then decelerates at 10 m/s². The section first calculates the distance covered during the driver's reaction time and the distance covered during deceleration to determine if the car will hit the barrier. The calculations are based on converting units to meters per second and applying the formula for deceleration to find the stopping distance.

05:06

🔢 Determining Maximum Safe Speed to Avoid Collision

This part of the script focuses on solving for the maximum speed at which the car could travel without hitting the barrier. It involves setting up an equation where the sum of the reaction distance and the deceleration distance equals 40 meters. The formula for final velocity (V² = u² + 2as) is rearranged to solve for the initial velocity (u), resulting in a quadratic equation. The solution to this equation gives the maximum speed, which is found to be 14.26 m/s, indicating that at this speed, the car would just avoid hitting the barrier.

10:12

📚 Conclusion and Encouragement to Practice

The final paragraph wraps up the physics problem by emphasizing the importance of practice and understanding the problem-solving process. It highlights the significance of visualizing the scenario and breaking down the problem into manageable parts. The presenter encourages viewers to ask questions if they have any difficulties and ends with a call to like, subscribe, and continue practicing physics for mastery.

Keywords

💡Physics

Physics is a branch of science that deals with the study of matter and energy, including their motion, behavior, and interactions. In the video, physics is the main theme as it explores the principles of motion and deceleration in the context of a car's stopping distance and speed.

💡Deceleration

Deceleration is the rate at which something slows down, typically measured in meters per second squared (m/s^2). The video uses deceleration to calculate how quickly the car can come to a stop, with a deceleration rate of 10 m/s^2 being a key factor in determining whether the car hits the barrier.

💡Reaction Time

Reaction time refers to the period it takes for a person to respond to a stimulus. In the script, the driver's reaction time of 0.75 seconds is crucial in calculating the distance covered before the brakes are applied, impacting the outcome of the scenario.

💡Stopping Distance

Stopping distance is the distance required for a moving vehicle to come to a complete stop. The video breaks down the stopping distance into two parts: the distance covered during reaction time and the distance covered during deceleration.

💡Barrier

In the context of the video, a barrier represents an obstacle that the car must avoid. The script discusses whether the car, traveling at a certain speed, can stop before hitting a barrier 40 meters away.

💡Velocity

Velocity is the speed of an object in a specific direction. The video starts with the car traveling at a velocity of 90 km/h, which is then converted to meters per second (25 m/s) for calculations.

💡Acceleration

Acceleration is the rate of change of velocity. While the video primarily discusses deceleration, the concept of acceleration is inversely related to it, as the car slows down from its initial velocity to a stop.

💡Quadratic Formula

The quadratic formula is used to solve quadratic equations of the form ax^2 + bx + c = 0. In the video, the quadratic formula is mentioned as a method to find the maximum speed at which the car could travel without hitting the barrier.

💡Meters per Second (m/s)

Meters per second is the standard unit of speed in the International System of Units (SI). The video converts the car's speed from kilometers per hour to meters per second to facilitate the physics calculations.

💡Maximum Speed

Maximum speed in this context refers to the highest velocity at which the car can travel without colliding with the barrier, given the driver's reaction time and the car's deceleration rate. The video calculates this speed as part of the problem-solving process.

💡Kinematic Equations

Kinematic equations relate the motion of an object to its initial and final velocities, acceleration, and time. The video uses the specific kinematic equation V^2 = U^2 + 2as to determine the stopping distance and maximum safe speed.

Highlights

The physics problem involves a car traveling at 90 km/h and the driver's reaction time to a barrier.

The car's speed is converted to meters per second for standard unit calculations.

The scenario is visualized to break down the problem into manageable parts.

The driver's reaction time is accounted for in the calculation of stopping distance.

The deceleration rate of the car is used to determine the stopping distance.

The problem is divided into two phases: reaction time and deceleration.

The formula V^2 = u^2 + 2as is used to calculate the stopping distance.

The car's initial speed and deceleration rate are used to find the stopping distance.

The total stopping distance is compared to the distance to the barrier to determine if the car will stop in time.

The car does not stop in time, indicating it will hit the barrier.

Part B of the problem explores the maximum safe speed to avoid hitting the barrier.

The quadratic formula is introduced to solve for the maximum safe speed.

The relationship between initial velocity, reaction time, and stopping distance is established.

A quadratic equation is set up to find the maximum initial velocity that allows stopping before the barrier.

The solution to the quadratic equation gives the maximum safe speed as 14.26 m/s.

The importance of practice in solving physics problems is emphasized.

The video concludes with an invitation for viewers to ask questions and engage with the content.

Transcripts

g'day guys got a physics question for

each day where we've got the driver of a

car doing 90 kilometres an hour

suddenly sees the lights of a barrier 40

metres ahead

it takes the drivers 0.75 seconds before

he applies the brakes and once he does

begin to break he decelerate at a rate

of 10 meters per second squared now the

two questions we've gotta answer are

does he hit the barrier very important

question and B which i think is the more

interesting question what would be the

maximum speed at which the car could

travel and not hit the barrier

40 metres ahead okay so first of all

when I'm whenever I'm doing a question

like this I like to sort of draw up a

picture of my scenario so I can sort of

visualize what I'm trying to deal with

so I'm gonna go ahead and just draw a

picture of my scenario okay so now we've

got a picture what we need to do is we

need to convert all of our figures into

like standard units so in here we have

90 kilometres an hour we have to convert

that into m/s so the way we do that is

we're just going to go 90 kilometers an

hour divided by 3.6 and that's equal to

25 meters per second great and this is

in meters that's in seconds that's in

meters per second squared

great so now we have everything in the

same standard units we can begin the

actual working out of the question so

Part A now does he hit the barrier so

what we first have to sort of

acknowledge here is the motion of the

guys in the car is going to be broken up

into two parts I think the best way to

describe it is there's the part where

they're still going whilst they're

reacting to the fact that they've just

seen this barrier and then there's the

part where they decelerate so what we're

gonna do is we're gonna break this 40

meters up into two components

this component and we've got this

component so the first component is

where they're going to be driving along

their merry way whilst they are still

reacting to the fact they've seen it and

jamming on the brake and then the second

components when they'll be decelerating

so let's call this one this is our

distance s1 and let's call this distance

s2 just because it's

it'll be easier to just break the whole

thing into two different parts now from

here what I'm going to do is I'm going

to go well the first part s1 is just

going to be how fast is going multiplied

by the time that II go

drives it for so the reaction times in

this case we're going to go s1 it's just

equal to velocity times time which in

this case is equal to 25 meters a second

times by 0.75 seconds we've done it is

there and that guy's is equal to

eighteen point seven five meters so

almost half of the stopping distance is

now taken up with him reacting to the

wall being there so now what we're going

to do is we're going to figure out well

how long is it going to take him to stop

from 25 meters a second to nothing and

for this part of the question we're

going to make use of the formula V

squared equals u squared plus 2 times a

times s and we're going to try and solve

for s now V squared is our final

velocity which hopefully will be

zero if we're able to stop in time and

that's going to equal au squared which

is you is 25 our initial velocity 25

squared plus two times acceleration

now we're decelerating at 10 meters a

second so that's going to be negative 10

times our distance s and this is gonna

be s to cool so let's work out what we

have here we just take it up to here we

have 0 equals 25 squared is 625 minus 20

times s 2 cool so s 2 is going to be

equal to 625 divided by 20 which guys is

equal to thirty one point two five

meters now hopefully you guys will

realize that 18.75

plus thirty one point two five is

greater than 40 so then the answer for

Part A of this question is no he will

not stop in

time full-stop great now what we're

going to do now is we're gonna start

Part B now Part B I thought is quite a

lot more interesting than Part A because

we have to basically do exactly the same

thing but we have to do everything in

terms of his initial velocity and then

try and solve it so what we're going to

do is for Part B the way that we're

going to set this out is we're going

through the maximum speed he's going to

stop as he's touching the wall so he's

gonna nudge the wall or it's just gonna

stop a nanometer from the wall so as a

result we know that s1 plus s2 is has to

equal you guessed it forty meters okay

for s1 we know that s1 is equal to our

initial velocity which we're going to

label X because that's what we're trying

to solve for times the time in which we

take to react which is 0.75 so in this

case s1 is equal to 0.75 times the

velocity that we are going we're going

to try and solve for X the s2 is a

little bit more complicated so s2 is

we're gonna we used the formula V

squared equals u squared plus 2a s now

what we're all hopeful you guys remember

is V squared our final velocity was zero

when you stopped your final velocity

will obviously be zero u squared is what

we're trying to find which we're going

to call x squared plus two times the

acceleration which is negative ten

comes by the distance now this part is a

clever bit we'll do that in the second

bit so the distance we'll call it that s

too great so as you can see at the

moment guys we've got two variables in

this second equation let's just make

everything in terms of s2 so we're gonna

have s2 Jesus

s2 is equal to x squared on 20 great now

from here we have to have something that

relates s1 to s2 so we can get rid of

one of these variables but if we look

what we could do is rather than even

doing that we can just plug in our s

ones and our s2 s into this initial

function so stay with me guys this s1

plus s2 equals 44 s1 what we're going to

do is we're going to write 0.75 X plus

s2 which is x squared on 20 and that's

going to equal 40 now what we can do is

we can make this all over the same base

because this is equal to 3 over 4 X so

we can multiply everything if we put

this all on the same base 3 over 4 will

become 15 over 20 so we're gonna have 15

X plus x squared all on 20 equals 40 and

then what we can do is we can take that

20 to the other side

by multiplying both sides by 20 and 40

times 20 is 800 and then what we're

gonna do is we're gonna bring that 800

over to the other side and we're gonna

have x squared plus 15 X minus 800

equals zero now from here you can either

solve it using the quadratic formula and

just for you know this is just off from

memory its X would equal the opposite of

B plus or minus the square root of b

squared minus 4ac all on 2a you could

use that you could use a calculator with

solving ability or you could complete

the square they'll all give you the same

answer and that is that X is going to

have to be equal to negative 15 plus or

minus 5 times the square root of 137

onto which one of these will give a

negative answer which doesn't make sense

but the one that gives a positive answer

gives us 14 point 2 6 meters per second

and that my friends is our the maximum

speed that this guy can be going so he

was going 25 initially and as a result

he hit the wall so if it was going 14

and a quarter of meters a second he

wouldn't have hit the wall well he would

just tap the wall I guess so

you know guys it's a relatively

complicated question which involves

breaking down a stopping distance into a

stopping distance that is

part of your reaction time and a

stopping distance which is associated to

your D acceleration but once you've sort

of separated them and I think the

picture is kind of vital in this case

you it makes the question a hell of a

lot easier but again practice practice

practice makes perfect if you have any

problems with it just ask me a question

I've no problem responding to your

comments but yeah just keep on

practicing guys give me a like subscribe

to my channel and enjoy your physics see

you next time