Determining distance between two distances given acceleration, max velocity, deceleration and total time.












1












$begingroup$


PROBLEM: A train accelerates from rest at one train station with $a= 0.72t$ $m/s^2$, until it reaches its maximum speed of $18 m/s$. Its deceleration heading into the next train station is $3.0 m/s^2$. If the journey between the two stations takes 6 minutes and 20 seconds, what is the distance between the stations?



WORKED ANSWER: I initially drew a $v-t$ graph, with maximum velocity of $18 m/s$ and maximum time of $380$ seconds. So the graph looks like a triangle with a peak of $18$. One edge of the triangle has a slope of $0.72t$ and the other edge has a slope of $-3$. So, the triangle is halved at time $t_1$ which I presumed is unknown.



PERSONAL QUESTION 1/2: According to the problem given, did I draw the graph correctly in order for me to answer the question?



WORKED ANSWER continued:



For $t le t_1$ $v=0.36t^2$:



$18 = 0.36t_1^2$



$t_1=5sqrt2$



$therefore$ $Delta x=A_1+A_2$



$=(1/2*5sqrt2*18)+(1/2*18*(380-5sqrt2))$
$=3420m$



PERSONAL QUESTION 2/2: If the graph I have drawn is correct, is my procedure correct and hence is my answer correct?






calculus arithmetic-dynamics







share|cite|improve this question









$endgroup$












  • $begingroup$
    You have not shown the graph. Yes, $t_1=5sqrt2$, and the graph is a parabola from $0to5sqrt2$, having the equation $v=0.36t^2$. This means the portion of the graph for $tle t_1$ is not a straight line, and your $A_1$ term is erroneous. $$A_1=int_0^{5sqrt2}0.36t^2dt$$
    $endgroup$
    – Shubham Johri
    Jan 10 at 7:28












  • $begingroup$
    @ShubhamJohri, so does $A_1+$$A_2=$$Delta x$ still apply but I just made a mistake for $A_1$, but $A_2$ is correct?
    $endgroup$
    – KPC2
    Jan 10 at 7:33












  • $begingroup$
    I believe the $t$ in $a=0.72t$ is in minutes
    $endgroup$
    – Shubham Johri
    Jan 10 at 7:41










  • $begingroup$
    Since the second part involves constant acceleration, your $A_2$ is correct.
    $endgroup$
    – KM101
    Jan 10 at 7:47










  • $begingroup$
    @KM101 So is $A_1 = integral$ correct? Wondering since I'm unsure.
    $endgroup$
    – KPC2
    Jan 10 at 7:54
















1












$begingroup$


PROBLEM: A train accelerates from rest at one train station with $a= 0.72t$ $m/s^2$, until it reaches its maximum speed of $18 m/s$. Its deceleration heading into the next train station is $3.0 m/s^2$. If the journey between the two stations takes 6 minutes and 20 seconds, what is the distance between the stations?



WORKED ANSWER: I initially drew a $v-t$ graph, with maximum velocity of $18 m/s$ and maximum time of $380$ seconds. So the graph looks like a triangle with a peak of $18$. One edge of the triangle has a slope of $0.72t$ and the other edge has a slope of $-3$. So, the triangle is halved at time $t_1$ which I presumed is unknown.



PERSONAL QUESTION 1/2: According to the problem given, did I draw the graph correctly in order for me to answer the question?



WORKED ANSWER continued:



For $t le t_1$ $v=0.36t^2$:



$18 = 0.36t_1^2$



$t_1=5sqrt2$



$therefore$ $Delta x=A_1+A_2$



$=(1/2*5sqrt2*18)+(1/2*18*(380-5sqrt2))$
$=3420m$



PERSONAL QUESTION 2/2: If the graph I have drawn is correct, is my procedure correct and hence is my answer correct?






calculus arithmetic-dynamics







share|cite|improve this question









$endgroup$












  • $begingroup$
    You have not shown the graph. Yes, $t_1=5sqrt2$, and the graph is a parabola from $0to5sqrt2$, having the equation $v=0.36t^2$. This means the portion of the graph for $tle t_1$ is not a straight line, and your $A_1$ term is erroneous. $$A_1=int_0^{5sqrt2}0.36t^2dt$$
    $endgroup$
    – Shubham Johri
    Jan 10 at 7:28












  • $begingroup$
    @ShubhamJohri, so does $A_1+$$A_2=$$Delta x$ still apply but I just made a mistake for $A_1$, but $A_2$ is correct?
    $endgroup$
    – KPC2
    Jan 10 at 7:33












  • $begingroup$
    I believe the $t$ in $a=0.72t$ is in minutes
    $endgroup$
    – Shubham Johri
    Jan 10 at 7:41










  • $begingroup$
    Since the second part involves constant acceleration, your $A_2$ is correct.
    $endgroup$
    – KM101
    Jan 10 at 7:47










  • $begingroup$
    @KM101 So is $A_1 = integral$ correct? Wondering since I'm unsure.
    $endgroup$
    – KPC2
    Jan 10 at 7:54














1












1








1





$begingroup$


PROBLEM: A train accelerates from rest at one train station with $a= 0.72t$ $m/s^2$, until it reaches its maximum speed of $18 m/s$. Its deceleration heading into the next train station is $3.0 m/s^2$. If the journey between the two stations takes 6 minutes and 20 seconds, what is the distance between the stations?



WORKED ANSWER: I initially drew a $v-t$ graph, with maximum velocity of $18 m/s$ and maximum time of $380$ seconds. So the graph looks like a triangle with a peak of $18$. One edge of the triangle has a slope of $0.72t$ and the other edge has a slope of $-3$. So, the triangle is halved at time $t_1$ which I presumed is unknown.



PERSONAL QUESTION 1/2: According to the problem given, did I draw the graph correctly in order for me to answer the question?



WORKED ANSWER continued:



For $t le t_1$ $v=0.36t^2$:



$18 = 0.36t_1^2$



$t_1=5sqrt2$



$therefore$ $Delta x=A_1+A_2$



$=(1/2*5sqrt2*18)+(1/2*18*(380-5sqrt2))$
$=3420m$



PERSONAL QUESTION 2/2: If the graph I have drawn is correct, is my procedure correct and hence is my answer correct?






calculus arithmetic-dynamics







share|cite|improve this question









$endgroup$




PROBLEM: A train accelerates from rest at one train station with $a= 0.72t$ $m/s^2$, until it reaches its maximum speed of $18 m/s$. Its deceleration heading into the next train station is $3.0 m/s^2$. If the journey between the two stations takes 6 minutes and 20 seconds, what is the distance between the stations?



WORKED ANSWER: I initially drew a $v-t$ graph, with maximum velocity of $18 m/s$ and maximum time of $380$ seconds. So the graph looks like a triangle with a peak of $18$. One edge of the triangle has a slope of $0.72t$ and the other edge has a slope of $-3$. So, the triangle is halved at time $t_1$ which I presumed is unknown.



PERSONAL QUESTION 1/2: According to the problem given, did I draw the graph correctly in order for me to answer the question?



WORKED ANSWER continued:



For $t le t_1$ $v=0.36t^2$:



$18 = 0.36t_1^2$



$t_1=5sqrt2$



$therefore$ $Delta x=A_1+A_2$



$=(1/2*5sqrt2*18)+(1/2*18*(380-5sqrt2))$
$=3420m$



PERSONAL QUESTION 2/2: If the graph I have drawn is correct, is my procedure correct and hence is my answer correct?






calculus arithmetic-dynamics




calculus arithmetic-dynamics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 10 at 7:01









KPC2KPC2

184




184












  • $begingroup$
    You have not shown the graph. Yes, $t_1=5sqrt2$, and the graph is a parabola from $0to5sqrt2$, having the equation $v=0.36t^2$. This means the portion of the graph for $tle t_1$ is not a straight line, and your $A_1$ term is erroneous. $$A_1=int_0^{5sqrt2}0.36t^2dt$$
    $endgroup$
    – Shubham Johri
    Jan 10 at 7:28












  • $begingroup$
    @ShubhamJohri, so does $A_1+$$A_2=$$Delta x$ still apply but I just made a mistake for $A_1$, but $A_2$ is correct?
    $endgroup$
    – KPC2
    Jan 10 at 7:33












  • $begingroup$
    I believe the $t$ in $a=0.72t$ is in minutes
    $endgroup$
    – Shubham Johri
    Jan 10 at 7:41










  • $begingroup$
    Since the second part involves constant acceleration, your $A_2$ is correct.
    $endgroup$
    – KM101
    Jan 10 at 7:47










  • $begingroup$
    @KM101 So is $A_1 = integral$ correct? Wondering since I'm unsure.
    $endgroup$
    – KPC2
    Jan 10 at 7:54


















  • $begingroup$
    You have not shown the graph. Yes, $t_1=5sqrt2$, and the graph is a parabola from $0to5sqrt2$, having the equation $v=0.36t^2$. This means the portion of the graph for $tle t_1$ is not a straight line, and your $A_1$ term is erroneous. $$A_1=int_0^{5sqrt2}0.36t^2dt$$
    $endgroup$
    – Shubham Johri
    Jan 10 at 7:28












  • $begingroup$
    @ShubhamJohri, so does $A_1+$$A_2=$$Delta x$ still apply but I just made a mistake for $A_1$, but $A_2$ is correct?
    $endgroup$
    – KPC2
    Jan 10 at 7:33












  • $begingroup$
    I believe the $t$ in $a=0.72t$ is in minutes
    $endgroup$
    – Shubham Johri
    Jan 10 at 7:41










  • $begingroup$
    Since the second part involves constant acceleration, your $A_2$ is correct.
    $endgroup$
    – KM101
    Jan 10 at 7:47










  • $begingroup$
    @KM101 So is $A_1 = integral$ correct? Wondering since I'm unsure.
    $endgroup$
    – KPC2
    Jan 10 at 7:54
















$begingroup$
You have not shown the graph. Yes, $t_1=5sqrt2$, and the graph is a parabola from $0to5sqrt2$, having the equation $v=0.36t^2$. This means the portion of the graph for $tle t_1$ is not a straight line, and your $A_1$ term is erroneous. $$A_1=int_0^{5sqrt2}0.36t^2dt$$
$endgroup$
– Shubham Johri
Jan 10 at 7:28






$begingroup$
You have not shown the graph. Yes, $t_1=5sqrt2$, and the graph is a parabola from $0to5sqrt2$, having the equation $v=0.36t^2$. This means the portion of the graph for $tle t_1$ is not a straight line, and your $A_1$ term is erroneous. $$A_1=int_0^{5sqrt2}0.36t^2dt$$
$endgroup$
– Shubham Johri
Jan 10 at 7:28














$begingroup$
@ShubhamJohri, so does $A_1+$$A_2=$$Delta x$ still apply but I just made a mistake for $A_1$, but $A_2$ is correct?
$endgroup$
– KPC2
Jan 10 at 7:33






$begingroup$
@ShubhamJohri, so does $A_1+$$A_2=$$Delta x$ still apply but I just made a mistake for $A_1$, but $A_2$ is correct?
$endgroup$
– KPC2
Jan 10 at 7:33














$begingroup$
I believe the $t$ in $a=0.72t$ is in minutes
$endgroup$
– Shubham Johri
Jan 10 at 7:41




$begingroup$
I believe the $t$ in $a=0.72t$ is in minutes
$endgroup$
– Shubham Johri
Jan 10 at 7:41












$begingroup$
Since the second part involves constant acceleration, your $A_2$ is correct.
$endgroup$
– KM101
Jan 10 at 7:47




$begingroup$
Since the second part involves constant acceleration, your $A_2$ is correct.
$endgroup$
– KM101
Jan 10 at 7:47












$begingroup$
@KM101 So is $A_1 = integral$ correct? Wondering since I'm unsure.
$endgroup$
– KPC2
Jan 10 at 7:54




$begingroup$
@KM101 So is $A_1 = integral$ correct? Wondering since I'm unsure.
$endgroup$
– KPC2
Jan 10 at 7:54










1 Answer
1






active

oldest

votes


















0












$begingroup$

Your graph for the first part isn’t correct. The integral of a linear function gives a parabola.



Just like how uniformly increasing velocity (constant acceleration) produces a parabolic $x-t$ (position-time) graph. For $a = 0.72t$, $j = 0.72$. $j$ represents “jerk” (rate of change of acceleration).



Hence, $$Delta x = frac{1}{6}jt^3+frac{1}{2}a_0t^2+v_0t = frac{1}{6}(0.72)left(5sqrt{2}right)^3 = 30sqrt{2}$$



(Since the object starts from rest, the terms involving $a_0$ and $v_0$ become $0$.) The formula above can be derived via repeated integration.



You can also use integration, which is also viable:



$$v(t) = int a(t)dt = int 0.72t = frac{1}{2}0.72t^2 = 0.36t^2$$



You have $t_0 = 0$ and $t_1 = 5sqrt{2}$, so



$$Delta x = int_{0}^{5sqrt{2}}0.36t^2 = frac{1}{3}(0.36)left(5sqrt{2}right)^3 = 30sqrt{2}$$



The rest of your answer is correct because there is constant acceleration, producing a triangular $v-t$ graph.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The acceleration is $0.72t$, not $0.72$
    $endgroup$
    – Shubham Johri
    Jan 10 at 7:25










  • $begingroup$
    Thanks for pointing out, I’ve edited the answer accordingly.
    $endgroup$
    – KM101
    Jan 10 at 7:43










  • $begingroup$
    Is this $Delta x$ for $A_1$?
    $endgroup$
    – KPC2
    Jan 10 at 7:56










  • $begingroup$
    Yes, your $A_2$ is correct.
    $endgroup$
    – KM101
    Jan 10 at 7:56










  • $begingroup$
    Also, sorry for the repetitive editing, I didn’t initially read the question correctly.
    $endgroup$
    – KM101
    Jan 10 at 8:01











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Your graph for the first part isn’t correct. The integral of a linear function gives a parabola.



Just like how uniformly increasing velocity (constant acceleration) produces a parabolic $x-t$ (position-time) graph. For $a = 0.72t$, $j = 0.72$. $j$ represents “jerk” (rate of change of acceleration).



Hence, $$Delta x = frac{1}{6}jt^3+frac{1}{2}a_0t^2+v_0t = frac{1}{6}(0.72)left(5sqrt{2}right)^3 = 30sqrt{2}$$



(Since the object starts from rest, the terms involving $a_0$ and $v_0$ become $0$.) The formula above can be derived via repeated integration.



You can also use integration, which is also viable:



$$v(t) = int a(t)dt = int 0.72t = frac{1}{2}0.72t^2 = 0.36t^2$$



You have $t_0 = 0$ and $t_1 = 5sqrt{2}$, so



$$Delta x = int_{0}^{5sqrt{2}}0.36t^2 = frac{1}{3}(0.36)left(5sqrt{2}right)^3 = 30sqrt{2}$$



The rest of your answer is correct because there is constant acceleration, producing a triangular $v-t$ graph.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The acceleration is $0.72t$, not $0.72$
    $endgroup$
    – Shubham Johri
    Jan 10 at 7:25










  • $begingroup$
    Thanks for pointing out, I’ve edited the answer accordingly.
    $endgroup$
    – KM101
    Jan 10 at 7:43










  • $begingroup$
    Is this $Delta x$ for $A_1$?
    $endgroup$
    – KPC2
    Jan 10 at 7:56










  • $begingroup$
    Yes, your $A_2$ is correct.
    $endgroup$
    – KM101
    Jan 10 at 7:56










  • $begingroup$
    Also, sorry for the repetitive editing, I didn’t initially read the question correctly.
    $endgroup$
    – KM101
    Jan 10 at 8:01
















0












$begingroup$

Your graph for the first part isn’t correct. The integral of a linear function gives a parabola.



Just like how uniformly increasing velocity (constant acceleration) produces a parabolic $x-t$ (position-time) graph. For $a = 0.72t$, $j = 0.72$. $j$ represents “jerk” (rate of change of acceleration).



Hence, $$Delta x = frac{1}{6}jt^3+frac{1}{2}a_0t^2+v_0t = frac{1}{6}(0.72)left(5sqrt{2}right)^3 = 30sqrt{2}$$



(Since the object starts from rest, the terms involving $a_0$ and $v_0$ become $0$.) The formula above can be derived via repeated integration.



You can also use integration, which is also viable:



$$v(t) = int a(t)dt = int 0.72t = frac{1}{2}0.72t^2 = 0.36t^2$$



You have $t_0 = 0$ and $t_1 = 5sqrt{2}$, so



$$Delta x = int_{0}^{5sqrt{2}}0.36t^2 = frac{1}{3}(0.36)left(5sqrt{2}right)^3 = 30sqrt{2}$$



The rest of your answer is correct because there is constant acceleration, producing a triangular $v-t$ graph.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The acceleration is $0.72t$, not $0.72$
    $endgroup$
    – Shubham Johri
    Jan 10 at 7:25










  • $begingroup$
    Thanks for pointing out, I’ve edited the answer accordingly.
    $endgroup$
    – KM101
    Jan 10 at 7:43










  • $begingroup$
    Is this $Delta x$ for $A_1$?
    $endgroup$
    – KPC2
    Jan 10 at 7:56










  • $begingroup$
    Yes, your $A_2$ is correct.
    $endgroup$
    – KM101
    Jan 10 at 7:56










  • $begingroup$
    Also, sorry for the repetitive editing, I didn’t initially read the question correctly.
    $endgroup$
    – KM101
    Jan 10 at 8:01














0












0








0





$begingroup$

Your graph for the first part isn’t correct. The integral of a linear function gives a parabola.



Just like how uniformly increasing velocity (constant acceleration) produces a parabolic $x-t$ (position-time) graph. For $a = 0.72t$, $j = 0.72$. $j$ represents “jerk” (rate of change of acceleration).



Hence, $$Delta x = frac{1}{6}jt^3+frac{1}{2}a_0t^2+v_0t = frac{1}{6}(0.72)left(5sqrt{2}right)^3 = 30sqrt{2}$$



(Since the object starts from rest, the terms involving $a_0$ and $v_0$ become $0$.) The formula above can be derived via repeated integration.



You can also use integration, which is also viable:



$$v(t) = int a(t)dt = int 0.72t = frac{1}{2}0.72t^2 = 0.36t^2$$



You have $t_0 = 0$ and $t_1 = 5sqrt{2}$, so



$$Delta x = int_{0}^{5sqrt{2}}0.36t^2 = frac{1}{3}(0.36)left(5sqrt{2}right)^3 = 30sqrt{2}$$



The rest of your answer is correct because there is constant acceleration, producing a triangular $v-t$ graph.






share|cite|improve this answer











$endgroup$



Your graph for the first part isn’t correct. The integral of a linear function gives a parabola.



Just like how uniformly increasing velocity (constant acceleration) produces a parabolic $x-t$ (position-time) graph. For $a = 0.72t$, $j = 0.72$. $j$ represents “jerk” (rate of change of acceleration).



Hence, $$Delta x = frac{1}{6}jt^3+frac{1}{2}a_0t^2+v_0t = frac{1}{6}(0.72)left(5sqrt{2}right)^3 = 30sqrt{2}$$



(Since the object starts from rest, the terms involving $a_0$ and $v_0$ become $0$.) The formula above can be derived via repeated integration.



You can also use integration, which is also viable:



$$v(t) = int a(t)dt = int 0.72t = frac{1}{2}0.72t^2 = 0.36t^2$$



You have $t_0 = 0$ and $t_1 = 5sqrt{2}$, so



$$Delta x = int_{0}^{5sqrt{2}}0.36t^2 = frac{1}{3}(0.36)left(5sqrt{2}right)^3 = 30sqrt{2}$$



The rest of your answer is correct because there is constant acceleration, producing a triangular $v-t$ graph.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 10 at 8:13

























answered Jan 10 at 7:20









KM101KM101

5,9131523




5,9131523












  • $begingroup$
    The acceleration is $0.72t$, not $0.72$
    $endgroup$
    – Shubham Johri
    Jan 10 at 7:25










  • $begingroup$
    Thanks for pointing out, I’ve edited the answer accordingly.
    $endgroup$
    – KM101
    Jan 10 at 7:43










  • $begingroup$
    Is this $Delta x$ for $A_1$?
    $endgroup$
    – KPC2
    Jan 10 at 7:56










  • $begingroup$
    Yes, your $A_2$ is correct.
    $endgroup$
    – KM101
    Jan 10 at 7:56










  • $begingroup$
    Also, sorry for the repetitive editing, I didn’t initially read the question correctly.
    $endgroup$
    – KM101
    Jan 10 at 8:01


















  • $begingroup$
    The acceleration is $0.72t$, not $0.72$
    $endgroup$
    – Shubham Johri
    Jan 10 at 7:25










  • $begingroup$
    Thanks for pointing out, I’ve edited the answer accordingly.
    $endgroup$
    – KM101
    Jan 10 at 7:43










  • $begingroup$
    Is this $Delta x$ for $A_1$?
    $endgroup$
    – KPC2
    Jan 10 at 7:56










  • $begingroup$
    Yes, your $A_2$ is correct.
    $endgroup$
    – KM101
    Jan 10 at 7:56










  • $begingroup$
    Also, sorry for the repetitive editing, I didn’t initially read the question correctly.
    $endgroup$
    – KM101
    Jan 10 at 8:01
















$begingroup$
The acceleration is $0.72t$, not $0.72$
$endgroup$
– Shubham Johri
Jan 10 at 7:25




$begingroup$
The acceleration is $0.72t$, not $0.72$
$endgroup$
– Shubham Johri
Jan 10 at 7:25












$begingroup$
Thanks for pointing out, I’ve edited the answer accordingly.
$endgroup$
– KM101
Jan 10 at 7:43




$begingroup$
Thanks for pointing out, I’ve edited the answer accordingly.
$endgroup$
– KM101
Jan 10 at 7:43












$begingroup$
Is this $Delta x$ for $A_1$?
$endgroup$
– KPC2
Jan 10 at 7:56




$begingroup$
Is this $Delta x$ for $A_1$?
$endgroup$
– KPC2
Jan 10 at 7:56












$begingroup$
Yes, your $A_2$ is correct.
$endgroup$
– KM101
Jan 10 at 7:56




$begingroup$
Yes, your $A_2$ is correct.
$endgroup$
– KM101
Jan 10 at 7:56












$begingroup$
Also, sorry for the repetitive editing, I didn’t initially read the question correctly.
$endgroup$
– KM101
Jan 10 at 8:01




$begingroup$
Also, sorry for the repetitive editing, I didn’t initially read the question correctly.
$endgroup$
– KM101
Jan 10 at 8:01


















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