PHYSICS - Calculating the Braking Distance of a Car (Exam Question Example)
Summary
TLDRIn 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
🚗 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.
🔢 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.
📚 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.
Mindmap
Keywords
💡Physics
💡Deceleration
💡Reaction Time
💡Stopping Distance
💡Barrier
💡Velocity
💡Acceleration
💡Quadratic Formula
💡Meters per Second (m/s)
💡Maximum Speed
💡Kinematic Equations
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
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