Question about the braking distance
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Imagine a car travelling on a straight road at speed $u$ metres per second. The driver sees a kangaroo ahead and brakes to stop, with a reaction time of two seconds. In such circumstances, the distance covered $x$ seconds after seeing the kangaroo is given by:
$$d(x) =
begincases
ux, & textif $0 leq x leq 2$,\
ux - 2(x - 2)^2, & textif $2 leq x leq 2 + fracu4$.
endcases
$$
Let $f: [0,infty) to [0,infty)$ be such that $f(u)$ is the total braking distance before the car comes to a complete standstill corresponding to the initial speed $u$. Find a formula for $f$.
I've tried but just have no ideas what I have to do. Can anyone help me please!!
functions
asked Aug 12 at 8:27
Mike LoongBoong
63
add a comment |Â
up vote
-1
down vote
favorite
Imagine a car travelling on a straight road at speed $u$ metres per second. The driver sees a kangaroo ahead and brakes to stop, with a reaction time of two seconds. In such circumstances, the distance covered $x$ seconds after seeing the kangaroo is given by:
$$d(x) =
begincases
ux, & textif $0 leq x leq 2$,\
ux - 2(x - 2)^2, & textif $2 leq x leq 2 + fracu4$.
endcases
$$
Let $f: [0,infty) to [0,infty)$ be such that $f(u)$ is the total braking distance before the car comes to a complete standstill corresponding to the initial speed $u$. Find a formula for $f$.
I've tried but just have no ideas what I have to do. Can anyone help me please!!
functions
asked Aug 12 at 8:27
Mike LoongBoong
63
1
I think your first $f(x)$ should have a different name such as $d(x)$. Otherwise you are using $f$ as the name of two completely different functions, which is confusing. I have edited the question.
– gandalf61
Aug 12 at 8:36
Oh right! Thank you.
– Mike LoongBoong
Aug 12 at 8:37
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Imagine a car travelling on a straight road at speed $u$ metres per second. The driver sees a kangaroo ahead and brakes to stop, with a reaction time of two seconds. In such circumstances, the distance covered $x$ seconds after seeing the kangaroo is given by:
$$d(x) =
begincases
ux, & textif $0 leq x leq 2$,\
ux - 2(x - 2)^2, & textif $2 leq x leq 2 + fracu4$.
endcases
$$
Let $f: [0,infty) to [0,infty)$ be such that $f(u)$ is the total braking distance before the car comes to a complete standstill corresponding to the initial speed $u$. Find a formula for $f$.
I've tried but just have no ideas what I have to do. Can anyone help me please!!
functions
asked Aug 12 at 8:27
Mike LoongBoong
63
Imagine a car travelling on a straight road at speed $u$ metres per second. The driver sees a kangaroo ahead and brakes to stop, with a reaction time of two seconds. In such circumstances, the distance covered $x$ seconds after seeing the kangaroo is given by:
$$d(x) =
begincases
ux, & textif $0 leq x leq 2$,\
ux - 2(x - 2)^2, & textif $2 leq x leq 2 + fracu4$.
endcases
$$
Let $f: [0,infty) to [0,infty)$ be such that $f(u)$ is the total braking distance before the car comes to a complete standstill corresponding to the initial speed $u$. Find a formula for $f$.
I've tried but just have no ideas what I have to do. Can anyone help me please!!
functions
asked Aug 12 at 8:27
Mike LoongBoong
63
edited Aug 12 at 8:37
asked Aug 12 at 8:27
Mike LoongBoong
63
asked Aug 12 at 8:27
Mike LoongBoong
63
asked Aug 12 at 8:27
Mike LoongBoong
63
63
1
I think your first $f(x)$ should have a different name such as $d(x)$. Otherwise you are using $f$ as the name of two completely different functions, which is confusing. I have edited the question.
– gandalf61
Aug 12 at 8:36
Oh right! Thank you.
– Mike LoongBoong
Aug 12 at 8:37
add a comment |Â
1
I think your first $f(x)$ should have a different name such as $d(x)$. Otherwise you are using $f$ as the name of two completely different functions, which is confusing. I have edited the question.
– gandalf61
Aug 12 at 8:36
Oh right! Thank you.
– Mike LoongBoong
Aug 12 at 8:37
1
1
I think your first $f(x)$ should have a different name such as $d(x)$. Otherwise you are using $f$ as the name of two completely different functions, which is confusing. I have edited the question.
– gandalf61
Aug 12 at 8:36
I think your first $f(x)$ should have a different name such as $d(x)$. Otherwise you are using $f$ as the name of two completely different functions, which is confusing. I have edited the question.
– gandalf61
Aug 12 at 8:36
Oh right! Thank you.
– Mike LoongBoong
Aug 12 at 8:37
Oh right! Thank you.
– Mike LoongBoong
Aug 12 at 8:37
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
The car continues at speed $u$ for 2 seconds and then decelerates at a rate of $4$ metres per second per second for $fracu4$ seconds until it comes to rest.
Use the stopping time $x=2+fracu4$ in the formula you are given for $d(x)$ to find the stopping distance $f(u)$.
answered Aug 12 at 8:34
gandalf61
5,796522
Thank you but I wonder why the car decelerates at a rate of 4 m/s?
– Mike LoongBoong
Aug 12 at 8:40
Find the second derivative of $d(x)$ during the period after $x=2$.
– gandalf61
Aug 12 at 8:47
Sorry but I still don't understand though I thought about it through the last night. I have 3 questions: (1) Why we need to calculate the second derivative, (2) Why we can use the stopping time x = 2 + u/4 to calculate the stopping distance (And I think if we can do this, the problem is solved), (3) I think the braking distance (if including the reaction distance) is equal to d(2 + u/4) if 2 <=x <= 2 + u/4. Could you please explain for me?
– Mike LoongBoong
Aug 13 at 0:35
add a comment |Â
up vote
0
down vote
You have been given the position $d$ in terms of time (which is called $x$ here) so you can calculate the velocity by taking the derivative.
The car has stopped once it reached velocity $v= d'(x) = 0$.
answered Aug 12 at 9:28
mvw
30.7k22251
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The car continues at speed $u$ for 2 seconds and then decelerates at a rate of $4$ metres per second per second for $fracu4$ seconds until it comes to rest.
Use the stopping time $x=2+fracu4$ in the formula you are given for $d(x)$ to find the stopping distance $f(u)$.
answered Aug 12 at 8:34
gandalf61
5,796522
Thank you but I wonder why the car decelerates at a rate of 4 m/s?
– Mike LoongBoong
Aug 12 at 8:40
Find the second derivative of $d(x)$ during the period after $x=2$.
– gandalf61
Aug 12 at 8:47
Sorry but I still don't understand though I thought about it through the last night. I have 3 questions: (1) Why we need to calculate the second derivative, (2) Why we can use the stopping time x = 2 + u/4 to calculate the stopping distance (And I think if we can do this, the problem is solved), (3) I think the braking distance (if including the reaction distance) is equal to d(2 + u/4) if 2 <=x <= 2 + u/4. Could you please explain for me?
– Mike LoongBoong
Aug 13 at 0:35
add a comment |Â
up vote
1
down vote
The car continues at speed $u$ for 2 seconds and then decelerates at a rate of $4$ metres per second per second for $fracu4$ seconds until it comes to rest.
Use the stopping time $x=2+fracu4$ in the formula you are given for $d(x)$ to find the stopping distance $f(u)$.
answered Aug 12 at 8:34
gandalf61
5,796522
Thank you but I wonder why the car decelerates at a rate of 4 m/s?
– Mike LoongBoong
Aug 12 at 8:40
Find the second derivative of $d(x)$ during the period after $x=2$.
– gandalf61
Aug 12 at 8:47
Sorry but I still don't understand though I thought about it through the last night. I have 3 questions: (1) Why we need to calculate the second derivative, (2) Why we can use the stopping time x = 2 + u/4 to calculate the stopping distance (And I think if we can do this, the problem is solved), (3) I think the braking distance (if including the reaction distance) is equal to d(2 + u/4) if 2 <=x <= 2 + u/4. Could you please explain for me?
– Mike LoongBoong
Aug 13 at 0:35
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The car continues at speed $u$ for 2 seconds and then decelerates at a rate of $4$ metres per second per second for $fracu4$ seconds until it comes to rest.
Use the stopping time $x=2+fracu4$ in the formula you are given for $d(x)$ to find the stopping distance $f(u)$.
answered Aug 12 at 8:34
gandalf61
5,796522
The car continues at speed $u$ for 2 seconds and then decelerates at a rate of $4$ metres per second per second for $fracu4$ seconds until it comes to rest.
Use the stopping time $x=2+fracu4$ in the formula you are given for $d(x)$ to find the stopping distance $f(u)$.
answered Aug 12 at 8:34
gandalf61
5,796522
answered Aug 12 at 8:34
gandalf61
5,796522
answered Aug 12 at 8:34
gandalf61
5,796522
answered Aug 12 at 8:34
gandalf61
5,796522
5,796522
Thank you but I wonder why the car decelerates at a rate of 4 m/s?
– Mike LoongBoong
Aug 12 at 8:40
Find the second derivative of $d(x)$ during the period after $x=2$.
– gandalf61
Aug 12 at 8:47
Sorry but I still don't understand though I thought about it through the last night. I have 3 questions: (1) Why we need to calculate the second derivative, (2) Why we can use the stopping time x = 2 + u/4 to calculate the stopping distance (And I think if we can do this, the problem is solved), (3) I think the braking distance (if including the reaction distance) is equal to d(2 + u/4) if 2 <=x <= 2 + u/4. Could you please explain for me?
– Mike LoongBoong
Aug 13 at 0:35
add a comment |Â
Thank you but I wonder why the car decelerates at a rate of 4 m/s?
– Mike LoongBoong
Aug 12 at 8:40
Find the second derivative of $d(x)$ during the period after $x=2$.
– gandalf61
Aug 12 at 8:47
Sorry but I still don't understand though I thought about it through the last night. I have 3 questions: (1) Why we need to calculate the second derivative, (2) Why we can use the stopping time x = 2 + u/4 to calculate the stopping distance (And I think if we can do this, the problem is solved), (3) I think the braking distance (if including the reaction distance) is equal to d(2 + u/4) if 2 <=x <= 2 + u/4. Could you please explain for me?
– Mike LoongBoong
Aug 13 at 0:35
Thank you but I wonder why the car decelerates at a rate of 4 m/s?
– Mike LoongBoong
Aug 12 at 8:40
Thank you but I wonder why the car decelerates at a rate of 4 m/s?
– Mike LoongBoong
Aug 12 at 8:40
Find the second derivative of $d(x)$ during the period after $x=2$.
– gandalf61
Aug 12 at 8:47
Find the second derivative of $d(x)$ during the period after $x=2$.
– gandalf61
Aug 12 at 8:47
Sorry but I still don't understand though I thought about it through the last night. I have 3 questions: (1) Why we need to calculate the second derivative, (2) Why we can use the stopping time x = 2 + u/4 to calculate the stopping distance (And I think if we can do this, the problem is solved), (3) I think the braking distance (if including the reaction distance) is equal to d(2 + u/4) if 2 <=x <= 2 + u/4. Could you please explain for me?
– Mike LoongBoong
Aug 13 at 0:35
Sorry but I still don't understand though I thought about it through the last night. I have 3 questions: (1) Why we need to calculate the second derivative, (2) Why we can use the stopping time x = 2 + u/4 to calculate the stopping distance (And I think if we can do this, the problem is solved), (3) I think the braking distance (if including the reaction distance) is equal to d(2 + u/4) if 2 <=x <= 2 + u/4. Could you please explain for me?
– Mike LoongBoong
Aug 13 at 0:35
add a comment |Â
up vote
0
down vote
You have been given the position $d$ in terms of time (which is called $x$ here) so you can calculate the velocity by taking the derivative.
The car has stopped once it reached velocity $v= d'(x) = 0$.
answered Aug 12 at 9:28
mvw
30.7k22251
add a comment |Â
up vote
0
down vote
You have been given the position $d$ in terms of time (which is called $x$ here) so you can calculate the velocity by taking the derivative.
The car has stopped once it reached velocity $v= d'(x) = 0$.
answered Aug 12 at 9:28
mvw
30.7k22251
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You have been given the position $d$ in terms of time (which is called $x$ here) so you can calculate the velocity by taking the derivative.
The car has stopped once it reached velocity $v= d'(x) = 0$.
answered Aug 12 at 9:28
mvw
30.7k22251
You have been given the position $d$ in terms of time (which is called $x$ here) so you can calculate the velocity by taking the derivative.
The car has stopped once it reached velocity $v= d'(x) = 0$.
answered Aug 12 at 9:28
mvw
30.7k22251
answered Aug 12 at 9:28
mvw
30.7k22251
answered Aug 12 at 9:28
mvw
30.7k22251
answered Aug 12 at 9:28
mvw
30.7k22251
30.7k22251
add a comment |Â
add a comment |Â
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1
I think your first $f(x)$ should have a different name such as $d(x)$. Otherwise you are using $f$ as the name of two completely different functions, which is confusing. I have edited the question.
– gandalf61
Aug 12 at 8:36
Oh right! Thank you.
– Mike LoongBoong
Aug 12 at 8:37