What is the minimum distance from the car light in which he can bring his car to a stop?
$begingroup$
A motorist is traveling at a constant speed V0 and is approaching a traffic light. He wishes to stop the car at the light. However, his reaction time, that is, the time it takes him to put his foot on the brakes, is T and the maximum deceleration of the braked car is a. What is the minimum distance from the car light in which he can bring his car to a stop?
The answer is supposed to be (V0²/2a) + V0T but I don't know how to get there. What I did until now was that:
a = -a
v = -at + C
V0 = -aT + C
V0 + aT = C
v = -at + V0 + aT
s = (-a/2)t² + V0t + aTt + C
0 = (-a/2)0² + V0.0 + At.0 + C
0 = C
s = (-a/2)t² + V0t + aTt
v = -at + V0 + aT
0 = -at + V0 + aT
V0 + aT = at
(V0 + aT)/a = t
Then substituting time in the "s = (-a/2)t² + V0t + aTt" equation gives us
(V0² + a²T² + 2aT)/2a
Did I do something wrong?
calculus
asked Dec 17 '18 at 16:58
IrlydontknowIrlydontknow
82
$endgroup$
add a comment |
$begingroup$
A motorist is traveling at a constant speed V0 and is approaching a traffic light. He wishes to stop the car at the light. However, his reaction time, that is, the time it takes him to put his foot on the brakes, is T and the maximum deceleration of the braked car is a. What is the minimum distance from the car light in which he can bring his car to a stop?
The answer is supposed to be (V0²/2a) + V0T but I don't know how to get there. What I did until now was that:
a = -a
v = -at + C
V0 = -aT + C
V0 + aT = C
v = -at + V0 + aT
s = (-a/2)t² + V0t + aTt + C
0 = (-a/2)0² + V0.0 + At.0 + C
0 = C
s = (-a/2)t² + V0t + aTt
v = -at + V0 + aT
0 = -at + V0 + aT
V0 + aT = at
(V0 + aT)/a = t
Then substituting time in the "s = (-a/2)t² + V0t + aTt" equation gives us
(V0² + a²T² + 2aT)/2a
Did I do something wrong?
calculus
asked Dec 17 '18 at 16:58
IrlydontknowIrlydontknow
82
$endgroup$
$begingroup$
$s = (-a/2)t² + V_0t + aTt$ Where's your constant of integration?
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:16
$begingroup$
Fixed, it was 0 so it doesn't change a thing and thank you btw.
$endgroup$
– Irlydontknow
Dec 17 '18 at 17:19
$begingroup$
Hint: that's not true. What time interval are you integrating over? Is the car's displacement 0 at the start of that interval?
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:24
$begingroup$
what does $t=0$ represents for you ?
$endgroup$
– G Cab
Dec 17 '18 at 17:25
$begingroup$
Well, I believe that t represents the time and yes the car displacement at the start of the interval 0 is 0. It has an initial velocity of V0 but at the interval 0 it hasn't moved yet.
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:20
add a comment |
$begingroup$
A motorist is traveling at a constant speed V0 and is approaching a traffic light. He wishes to stop the car at the light. However, his reaction time, that is, the time it takes him to put his foot on the brakes, is T and the maximum deceleration of the braked car is a. What is the minimum distance from the car light in which he can bring his car to a stop?
The answer is supposed to be (V0²/2a) + V0T but I don't know how to get there. What I did until now was that:
a = -a
v = -at + C
V0 = -aT + C
V0 + aT = C
v = -at + V0 + aT
s = (-a/2)t² + V0t + aTt + C
0 = (-a/2)0² + V0.0 + At.0 + C
0 = C
s = (-a/2)t² + V0t + aTt
v = -at + V0 + aT
0 = -at + V0 + aT
V0 + aT = at
(V0 + aT)/a = t
Then substituting time in the "s = (-a/2)t² + V0t + aTt" equation gives us
(V0² + a²T² + 2aT)/2a
Did I do something wrong?
calculus
asked Dec 17 '18 at 16:58
IrlydontknowIrlydontknow
82
$endgroup$
A motorist is traveling at a constant speed V0 and is approaching a traffic light. He wishes to stop the car at the light. However, his reaction time, that is, the time it takes him to put his foot on the brakes, is T and the maximum deceleration of the braked car is a. What is the minimum distance from the car light in which he can bring his car to a stop?
The answer is supposed to be (V0²/2a) + V0T but I don't know how to get there. What I did until now was that:
a = -a
v = -at + C
V0 = -aT + C
V0 + aT = C
v = -at + V0 + aT
s = (-a/2)t² + V0t + aTt + C
0 = (-a/2)0² + V0.0 + At.0 + C
0 = C
s = (-a/2)t² + V0t + aTt
v = -at + V0 + aT
0 = -at + V0 + aT
V0 + aT = at
(V0 + aT)/a = t
Then substituting time in the "s = (-a/2)t² + V0t + aTt" equation gives us
(V0² + a²T² + 2aT)/2a
Did I do something wrong?
calculus
calculus
asked Dec 17 '18 at 16:58
IrlydontknowIrlydontknow
82
asked Dec 17 '18 at 16:58
IrlydontknowIrlydontknow
82
edited Dec 17 '18 at 17:20
Irlydontknow
asked Dec 17 '18 at 16:58
IrlydontknowIrlydontknow
82
asked Dec 17 '18 at 16:58
IrlydontknowIrlydontknow
82
asked Dec 17 '18 at 16:58
IrlydontknowIrlydontknow
82
82
$begingroup$
$s = (-a/2)t² + V_0t + aTt$ Where's your constant of integration?
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:16
$begingroup$
Fixed, it was 0 so it doesn't change a thing and thank you btw.
$endgroup$
– Irlydontknow
Dec 17 '18 at 17:19
$begingroup$
Hint: that's not true. What time interval are you integrating over? Is the car's displacement 0 at the start of that interval?
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:24
$begingroup$
what does $t=0$ represents for you ?
$endgroup$
– G Cab
Dec 17 '18 at 17:25
$begingroup$
Well, I believe that t represents the time and yes the car displacement at the start of the interval 0 is 0. It has an initial velocity of V0 but at the interval 0 it hasn't moved yet.
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:20
add a comment |
$begingroup$
$s = (-a/2)t² + V_0t + aTt$ Where's your constant of integration?
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:16
$begingroup$
Fixed, it was 0 so it doesn't change a thing and thank you btw.
$endgroup$
– Irlydontknow
Dec 17 '18 at 17:19
$begingroup$
Hint: that's not true. What time interval are you integrating over? Is the car's displacement 0 at the start of that interval?
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:24
$begingroup$
what does $t=0$ represents for you ?
$endgroup$
– G Cab
Dec 17 '18 at 17:25
$begingroup$
Well, I believe that t represents the time and yes the car displacement at the start of the interval 0 is 0. It has an initial velocity of V0 but at the interval 0 it hasn't moved yet.
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:20
$begingroup$
$s = (-a/2)t² + V_0t + aTt$ Where's your constant of integration?
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:16
$begingroup$
$s = (-a/2)t² + V_0t + aTt$ Where's your constant of integration?
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:16
$begingroup$
Fixed, it was 0 so it doesn't change a thing and thank you btw.
$endgroup$
– Irlydontknow
Dec 17 '18 at 17:19
$begingroup$
Fixed, it was 0 so it doesn't change a thing and thank you btw.
$endgroup$
– Irlydontknow
Dec 17 '18 at 17:19
$begingroup$
Hint: that's not true. What time interval are you integrating over? Is the car's displacement 0 at the start of that interval?
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:24
$begingroup$
Hint: that's not true. What time interval are you integrating over? Is the car's displacement 0 at the start of that interval?
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:24
$begingroup$
what does $t=0$ represents for you ?
$endgroup$
– G Cab
Dec 17 '18 at 17:25
$begingroup$
what does $t=0$ represents for you ?
$endgroup$
– G Cab
Dec 17 '18 at 17:25
$begingroup$
Well, I believe that t represents the time and yes the car displacement at the start of the interval 0 is 0. It has an initial velocity of V0 but at the interval 0 it hasn't moved yet.
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:20
$begingroup$
Well, I believe that t represents the time and yes the car displacement at the start of the interval 0 is 0. It has an initial velocity of V0 but at the interval 0 it hasn't moved yet.
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Following your scheme and indicating with$T$ the reaction time
and with $Delta t$ the following time interval to stop
$$
eqalign{
& a = - a cr
& v(t) = - at + C cr
& v(0) = V_0 = C cr
& v(t) = V_0 - at cr
& v(Delta t) = 0 = V_0 - aDelta t cr
& Delta t = V_0 /a cr
& s = V_0 T + overline {v(t)} ;Delta t = V_0 T + {{left( {v(0) + v(Delta t)} right)} over 2};Delta t = cr
& = V_0 T + {{left( {V_0 + 0} right)} over 2};{{V_0 } over a} = V_0 T + {1 over 2};{{V_0 ^{,2} } over a} cr}
$$
answered Dec 17 '18 at 17:25
G CabG Cab
19.4k31238
$endgroup$
$begingroup$
Thank you, I was able to see my mistake but I didn't understand a few things. Why did you sum V0T to the displacement function? Where did that V0T actually come from?
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:39
$begingroup$
until the driver actually press the brake, the car will continue with the constant initial speed (for a time T $to s_1=V_0T$). After that the deceleration will decrease (approx. linearly) the speed from $V_0$ to $0$ taking in $Delta t = V_0/a$. During this time the travelled distance $s_2$ will be equal to the average speed $V_0/2$ times $Delta t$.
$endgroup$
– G Cab
Dec 17 '18 at 23:10
$begingroup$
Ooh, now everything makes sense. Thank you!
$endgroup$
– Irlydontknow
Dec 18 '18 at 13:56
add a comment |
$begingroup$
$$v^2-u^2=2astag{Third equation of motion}$$
Put $v=0$(final velocity), $u=v_0$(initial velocity) and $a=-a$(deceleration) for this case to get$$v_0^2=2as$$
where $s$ is the distance further travelled.
answered Dec 17 '18 at 17:22
Sameer BahetiSameer Baheti
5168
$endgroup$
$begingroup$
Thank you, that's it. But I wanted to solve it with the antidifferentiation method.
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:41
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044173%2fwhat-is-the-minimum-distance-from-the-car-light-in-which-he-can-bring-his-car-to%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Following your scheme and indicating with$T$ the reaction time
and with $Delta t$ the following time interval to stop
$$
eqalign{
& a = - a cr
& v(t) = - at + C cr
& v(0) = V_0 = C cr
& v(t) = V_0 - at cr
& v(Delta t) = 0 = V_0 - aDelta t cr
& Delta t = V_0 /a cr
& s = V_0 T + overline {v(t)} ;Delta t = V_0 T + {{left( {v(0) + v(Delta t)} right)} over 2};Delta t = cr
& = V_0 T + {{left( {V_0 + 0} right)} over 2};{{V_0 } over a} = V_0 T + {1 over 2};{{V_0 ^{,2} } over a} cr}
$$
answered Dec 17 '18 at 17:25
G CabG Cab
19.4k31238
$endgroup$
$begingroup$
Thank you, I was able to see my mistake but I didn't understand a few things. Why did you sum V0T to the displacement function? Where did that V0T actually come from?
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:39
$begingroup$
until the driver actually press the brake, the car will continue with the constant initial speed (for a time T $to s_1=V_0T$). After that the deceleration will decrease (approx. linearly) the speed from $V_0$ to $0$ taking in $Delta t = V_0/a$. During this time the travelled distance $s_2$ will be equal to the average speed $V_0/2$ times $Delta t$.
$endgroup$
– G Cab
Dec 17 '18 at 23:10
$begingroup$
Ooh, now everything makes sense. Thank you!
$endgroup$
– Irlydontknow
Dec 18 '18 at 13:56
add a comment |
$begingroup$
Following your scheme and indicating with$T$ the reaction time
and with $Delta t$ the following time interval to stop
$$
eqalign{
& a = - a cr
& v(t) = - at + C cr
& v(0) = V_0 = C cr
& v(t) = V_0 - at cr
& v(Delta t) = 0 = V_0 - aDelta t cr
& Delta t = V_0 /a cr
& s = V_0 T + overline {v(t)} ;Delta t = V_0 T + {{left( {v(0) + v(Delta t)} right)} over 2};Delta t = cr
& = V_0 T + {{left( {V_0 + 0} right)} over 2};{{V_0 } over a} = V_0 T + {1 over 2};{{V_0 ^{,2} } over a} cr}
$$
answered Dec 17 '18 at 17:25
G CabG Cab
19.4k31238
$endgroup$
$begingroup$
Thank you, I was able to see my mistake but I didn't understand a few things. Why did you sum V0T to the displacement function? Where did that V0T actually come from?
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:39
$begingroup$
until the driver actually press the brake, the car will continue with the constant initial speed (for a time T $to s_1=V_0T$). After that the deceleration will decrease (approx. linearly) the speed from $V_0$ to $0$ taking in $Delta t = V_0/a$. During this time the travelled distance $s_2$ will be equal to the average speed $V_0/2$ times $Delta t$.
$endgroup$
– G Cab
Dec 17 '18 at 23:10
$begingroup$
Ooh, now everything makes sense. Thank you!
$endgroup$
– Irlydontknow
Dec 18 '18 at 13:56
add a comment |
$begingroup$
Following your scheme and indicating with$T$ the reaction time
and with $Delta t$ the following time interval to stop
$$
eqalign{
& a = - a cr
& v(t) = - at + C cr
& v(0) = V_0 = C cr
& v(t) = V_0 - at cr
& v(Delta t) = 0 = V_0 - aDelta t cr
& Delta t = V_0 /a cr
& s = V_0 T + overline {v(t)} ;Delta t = V_0 T + {{left( {v(0) + v(Delta t)} right)} over 2};Delta t = cr
& = V_0 T + {{left( {V_0 + 0} right)} over 2};{{V_0 } over a} = V_0 T + {1 over 2};{{V_0 ^{,2} } over a} cr}
$$
answered Dec 17 '18 at 17:25
G CabG Cab
19.4k31238
$endgroup$
Following your scheme and indicating with$T$ the reaction time
and with $Delta t$ the following time interval to stop
$$
eqalign{
& a = - a cr
& v(t) = - at + C cr
& v(0) = V_0 = C cr
& v(t) = V_0 - at cr
& v(Delta t) = 0 = V_0 - aDelta t cr
& Delta t = V_0 /a cr
& s = V_0 T + overline {v(t)} ;Delta t = V_0 T + {{left( {v(0) + v(Delta t)} right)} over 2};Delta t = cr
& = V_0 T + {{left( {V_0 + 0} right)} over 2};{{V_0 } over a} = V_0 T + {1 over 2};{{V_0 ^{,2} } over a} cr}
$$
answered Dec 17 '18 at 17:25
G CabG Cab
19.4k31238
answered Dec 17 '18 at 17:25
G CabG Cab
19.4k31238
answered Dec 17 '18 at 17:25
G CabG Cab
19.4k31238
answered Dec 17 '18 at 17:25
G CabG Cab
19.4k31238
19.4k31238
$begingroup$
Thank you, I was able to see my mistake but I didn't understand a few things. Why did you sum V0T to the displacement function? Where did that V0T actually come from?
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:39
$begingroup$
until the driver actually press the brake, the car will continue with the constant initial speed (for a time T $to s_1=V_0T$). After that the deceleration will decrease (approx. linearly) the speed from $V_0$ to $0$ taking in $Delta t = V_0/a$. During this time the travelled distance $s_2$ will be equal to the average speed $V_0/2$ times $Delta t$.
$endgroup$
– G Cab
Dec 17 '18 at 23:10
$begingroup$
Ooh, now everything makes sense. Thank you!
$endgroup$
– Irlydontknow
Dec 18 '18 at 13:56
add a comment |
$begingroup$
Thank you, I was able to see my mistake but I didn't understand a few things. Why did you sum V0T to the displacement function? Where did that V0T actually come from?
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:39
$begingroup$
until the driver actually press the brake, the car will continue with the constant initial speed (for a time T $to s_1=V_0T$). After that the deceleration will decrease (approx. linearly) the speed from $V_0$ to $0$ taking in $Delta t = V_0/a$. During this time the travelled distance $s_2$ will be equal to the average speed $V_0/2$ times $Delta t$.
$endgroup$
– G Cab
Dec 17 '18 at 23:10
$begingroup$
Ooh, now everything makes sense. Thank you!
$endgroup$
– Irlydontknow
Dec 18 '18 at 13:56
$begingroup$
Thank you, I was able to see my mistake but I didn't understand a few things. Why did you sum V0T to the displacement function? Where did that V0T actually come from?
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:39
$begingroup$
Thank you, I was able to see my mistake but I didn't understand a few things. Why did you sum V0T to the displacement function? Where did that V0T actually come from?
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:39
$begingroup$
until the driver actually press the brake, the car will continue with the constant initial speed (for a time T $to s_1=V_0T$). After that the deceleration will decrease (approx. linearly) the speed from $V_0$ to $0$ taking in $Delta t = V_0/a$. During this time the travelled distance $s_2$ will be equal to the average speed $V_0/2$ times $Delta t$.
$endgroup$
– G Cab
Dec 17 '18 at 23:10
$begingroup$
until the driver actually press the brake, the car will continue with the constant initial speed (for a time T $to s_1=V_0T$). After that the deceleration will decrease (approx. linearly) the speed from $V_0$ to $0$ taking in $Delta t = V_0/a$. During this time the travelled distance $s_2$ will be equal to the average speed $V_0/2$ times $Delta t$.
$endgroup$
– G Cab
Dec 17 '18 at 23:10
$begingroup$
Ooh, now everything makes sense. Thank you!
$endgroup$
– Irlydontknow
Dec 18 '18 at 13:56
$begingroup$
Ooh, now everything makes sense. Thank you!
$endgroup$
– Irlydontknow
Dec 18 '18 at 13:56
add a comment |
$begingroup$
$$v^2-u^2=2astag{Third equation of motion}$$
Put $v=0$(final velocity), $u=v_0$(initial velocity) and $a=-a$(deceleration) for this case to get$$v_0^2=2as$$
where $s$ is the distance further travelled.
answered Dec 17 '18 at 17:22
Sameer BahetiSameer Baheti
5168
$endgroup$
$begingroup$
Thank you, that's it. But I wanted to solve it with the antidifferentiation method.
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:41
add a comment |
$begingroup$
$$v^2-u^2=2astag{Third equation of motion}$$
Put $v=0$(final velocity), $u=v_0$(initial velocity) and $a=-a$(deceleration) for this case to get$$v_0^2=2as$$
where $s$ is the distance further travelled.
answered Dec 17 '18 at 17:22
Sameer BahetiSameer Baheti
5168
$endgroup$
$begingroup$
Thank you, that's it. But I wanted to solve it with the antidifferentiation method.
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:41
add a comment |
$begingroup$
$$v^2-u^2=2astag{Third equation of motion}$$
Put $v=0$(final velocity), $u=v_0$(initial velocity) and $a=-a$(deceleration) for this case to get$$v_0^2=2as$$
where $s$ is the distance further travelled.
answered Dec 17 '18 at 17:22
Sameer BahetiSameer Baheti
5168
$endgroup$
$$v^2-u^2=2astag{Third equation of motion}$$
Put $v=0$(final velocity), $u=v_0$(initial velocity) and $a=-a$(deceleration) for this case to get$$v_0^2=2as$$
where $s$ is the distance further travelled.
answered Dec 17 '18 at 17:22
Sameer BahetiSameer Baheti
5168
answered Dec 17 '18 at 17:22
Sameer BahetiSameer Baheti
5168
answered Dec 17 '18 at 17:22
Sameer BahetiSameer Baheti
5168
answered Dec 17 '18 at 17:22
Sameer BahetiSameer Baheti
5168
5168
$begingroup$
Thank you, that's it. But I wanted to solve it with the antidifferentiation method.
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:41
add a comment |
$begingroup$
Thank you, that's it. But I wanted to solve it with the antidifferentiation method.
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:41
$begingroup$
Thank you, that's it. But I wanted to solve it with the antidifferentiation method.
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:41
$begingroup$
Thank you, that's it. But I wanted to solve it with the antidifferentiation method.
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:41
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044173%2fwhat-is-the-minimum-distance-from-the-car-light-in-which-he-can-bring-his-car-to%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
$s = (-a/2)t² + V_0t + aTt$ Where's your constant of integration?
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:16
$begingroup$
Fixed, it was 0 so it doesn't change a thing and thank you btw.
$endgroup$
– Irlydontknow
Dec 17 '18 at 17:19
$begingroup$
Hint: that's not true. What time interval are you integrating over? Is the car's displacement 0 at the start of that interval?
$endgroup$
– eyeballfrog
Dec 17 '18 at 17:24
$begingroup$
what does $t=0$ represents for you ?
$endgroup$
– G Cab
Dec 17 '18 at 17:25
$begingroup$
Well, I believe that t represents the time and yes the car displacement at the start of the interval 0 is 0. It has an initial velocity of V0 but at the interval 0 it hasn't moved yet.
$endgroup$
– Irlydontknow
Dec 17 '18 at 19:20