Finding the distance if speed is $v=frac{63}{200}tcos t$
$begingroup$
So I have the following exercise:
Student is running in the straight corridor with the spdeed $v = frac{63}{200}tcos t$ m/s where $t$ is the time that has passed since the student started running in seconds. Find the distance that student has traveled
during the period of time $ t in [6,11]$ , the distance form the starting point and average speed.
What I found out is that student is $-3.24$ m from the starting point, but I can't find the total distance that he has travelled, since I can't put that $frac{63}{200}tcos t =0$ and $int|frac{63}{200}tcos t |$ doesn't exist either.
And I also don't know the equation for finding the average speed
calculus definite-integrals physics
edited Dec 7 '18 at 18:09
Harry49
6,13931132
asked Dec 7 '18 at 12:47
Student123Student123
536
$endgroup$
add a comment |
$begingroup$
So I have the following exercise:
Student is running in the straight corridor with the spdeed $v = frac{63}{200}tcos t$ m/s where $t$ is the time that has passed since the student started running in seconds. Find the distance that student has traveled
during the period of time $ t in [6,11]$ , the distance form the starting point and average speed.
What I found out is that student is $-3.24$ m from the starting point, but I can't find the total distance that he has travelled, since I can't put that $frac{63}{200}tcos t =0$ and $int|frac{63}{200}tcos t |$ doesn't exist either.
And I also don't know the equation for finding the average speed
calculus definite-integrals physics
edited Dec 7 '18 at 18:09
Harry49
6,13931132
asked Dec 7 '18 at 12:47
Student123Student123
536
$endgroup$
$begingroup$
The average speed is the distance traveled divided by the time spent traveling. What do you mean the integral doesn't exist? Of course it does.
$endgroup$
– saulspatz
Dec 7 '18 at 14:13
add a comment |
$begingroup$
So I have the following exercise:
Student is running in the straight corridor with the spdeed $v = frac{63}{200}tcos t$ m/s where $t$ is the time that has passed since the student started running in seconds. Find the distance that student has traveled
during the period of time $ t in [6,11]$ , the distance form the starting point and average speed.
What I found out is that student is $-3.24$ m from the starting point, but I can't find the total distance that he has travelled, since I can't put that $frac{63}{200}tcos t =0$ and $int|frac{63}{200}tcos t |$ doesn't exist either.
And I also don't know the equation for finding the average speed
calculus definite-integrals physics
edited Dec 7 '18 at 18:09
Harry49
6,13931132
asked Dec 7 '18 at 12:47
Student123Student123
536
$endgroup$
So I have the following exercise:
Student is running in the straight corridor with the spdeed $v = frac{63}{200}tcos t$ m/s where $t$ is the time that has passed since the student started running in seconds. Find the distance that student has traveled
during the period of time $ t in [6,11]$ , the distance form the starting point and average speed.
What I found out is that student is $-3.24$ m from the starting point, but I can't find the total distance that he has travelled, since I can't put that $frac{63}{200}tcos t =0$ and $int|frac{63}{200}tcos t |$ doesn't exist either.
And I also don't know the equation for finding the average speed
calculus definite-integrals physics
calculus definite-integrals physics
edited Dec 7 '18 at 18:09
Harry49
6,13931132
asked Dec 7 '18 at 12:47
Student123Student123
536
edited Dec 7 '18 at 18:09
Harry49
6,13931132
asked Dec 7 '18 at 12:47
Student123Student123
536
edited Dec 7 '18 at 18:09
Harry49
6,13931132
edited Dec 7 '18 at 18:09
Harry49
6,13931132
edited Dec 7 '18 at 18:09
Harry49
6,13931132
6,13931132
asked Dec 7 '18 at 12:47
Student123Student123
536
asked Dec 7 '18 at 12:47
Student123Student123
536
asked Dec 7 '18 at 12:47
Student123Student123
536
536
$begingroup$
The average speed is the distance traveled divided by the time spent traveling. What do you mean the integral doesn't exist? Of course it does.
$endgroup$
– saulspatz
Dec 7 '18 at 14:13
add a comment |
$begingroup$
The average speed is the distance traveled divided by the time spent traveling. What do you mean the integral doesn't exist? Of course it does.
$endgroup$
– saulspatz
Dec 7 '18 at 14:13
$begingroup$
The average speed is the distance traveled divided by the time spent traveling. What do you mean the integral doesn't exist? Of course it does.
$endgroup$
– saulspatz
Dec 7 '18 at 14:13
$begingroup$
The average speed is the distance traveled divided by the time spent traveling. What do you mean the integral doesn't exist? Of course it does.
$endgroup$
– saulspatz
Dec 7 '18 at 14:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The average of a function $f:mathbb{R}tomathbb{R}$ on $[a,b]subseteqtext{dom}(f)$ is
$$bar f_{[a,b]} = frac{1}{b-a}int^b_af$$
And by definition,
$$v=dot{x}$$
So total distance travelled will be
$$d=int_a^b|v|mathrm{d}t$$
It's not so hard to find the roots of your velocity function, and based on that, you can take apart the integral into $3$ smaller one, and remove the absolute value sign on them accordingly to the sign of the velocity.
answered Dec 7 '18 at 14:26
BotondBotond
5,6982732
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The average of a function $f:mathbb{R}tomathbb{R}$ on $[a,b]subseteqtext{dom}(f)$ is
$$bar f_{[a,b]} = frac{1}{b-a}int^b_af$$
And by definition,
$$v=dot{x}$$
So total distance travelled will be
$$d=int_a^b|v|mathrm{d}t$$
It's not so hard to find the roots of your velocity function, and based on that, you can take apart the integral into $3$ smaller one, and remove the absolute value sign on them accordingly to the sign of the velocity.
answered Dec 7 '18 at 14:26
BotondBotond
5,6982732
$endgroup$
add a comment |
$begingroup$
The average of a function $f:mathbb{R}tomathbb{R}$ on $[a,b]subseteqtext{dom}(f)$ is
$$bar f_{[a,b]} = frac{1}{b-a}int^b_af$$
And by definition,
$$v=dot{x}$$
So total distance travelled will be
$$d=int_a^b|v|mathrm{d}t$$
It's not so hard to find the roots of your velocity function, and based on that, you can take apart the integral into $3$ smaller one, and remove the absolute value sign on them accordingly to the sign of the velocity.
answered Dec 7 '18 at 14:26
BotondBotond
5,6982732
$endgroup$
add a comment |
$begingroup$
The average of a function $f:mathbb{R}tomathbb{R}$ on $[a,b]subseteqtext{dom}(f)$ is
$$bar f_{[a,b]} = frac{1}{b-a}int^b_af$$
And by definition,
$$v=dot{x}$$
So total distance travelled will be
$$d=int_a^b|v|mathrm{d}t$$
It's not so hard to find the roots of your velocity function, and based on that, you can take apart the integral into $3$ smaller one, and remove the absolute value sign on them accordingly to the sign of the velocity.
answered Dec 7 '18 at 14:26
BotondBotond
5,6982732
$endgroup$
The average of a function $f:mathbb{R}tomathbb{R}$ on $[a,b]subseteqtext{dom}(f)$ is
$$bar f_{[a,b]} = frac{1}{b-a}int^b_af$$
And by definition,
$$v=dot{x}$$
So total distance travelled will be
$$d=int_a^b|v|mathrm{d}t$$
It's not so hard to find the roots of your velocity function, and based on that, you can take apart the integral into $3$ smaller one, and remove the absolute value sign on them accordingly to the sign of the velocity.
answered Dec 7 '18 at 14:26
BotondBotond
5,6982732
answered Dec 7 '18 at 14:26
BotondBotond
5,6982732
answered Dec 7 '18 at 14:26
BotondBotond
5,6982732
answered Dec 7 '18 at 14:26
BotondBotond
5,6982732
5,6982732
add a comment |
add a comment |
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$begingroup$
The average speed is the distance traveled divided by the time spent traveling. What do you mean the integral doesn't exist? Of course it does.
$endgroup$
– saulspatz
Dec 7 '18 at 14:13