One question: I guess it's to do with suvat equations:Kinematics [closed]
$begingroup$
A car moves along a straight horizontal road from a point A to a point B, where AB = 885 m. The car accelerates from rest at A to a speed of 15 m s −1 at a constant rate a m s−2. The time for which the car accelerates is T seconds. The car maintains the speed of 15 m s−1 for T seconds. The car then decelerates at a constant rate of 2.5 m s−2 stopping at B.
Find the time for which the car decelerates.
How do you approach this question?
Other details:
Here's kinematic equations for distance and speed:
$x(t)=v_0*t pm frac{a*t^2}{2} tag{1}$
$v(t)= v_0 pm a*t tag{2}$
kinematics
edited Dec 14 '18 at 16:53
peterh
2,17351731
asked Dec 11 '18 at 20:54
Yusuf123Yusuf123
111
$endgroup$
closed as off-topic by Davide Giraudo, Dando18, John B, KReiser, platty Dec 12 '18 at 0:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, Dando18, John B, KReiser, platty
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
A car moves along a straight horizontal road from a point A to a point B, where AB = 885 m. The car accelerates from rest at A to a speed of 15 m s −1 at a constant rate a m s−2. The time for which the car accelerates is T seconds. The car maintains the speed of 15 m s−1 for T seconds. The car then decelerates at a constant rate of 2.5 m s−2 stopping at B.
Find the time for which the car decelerates.
How do you approach this question?
Other details:
Here's kinematic equations for distance and speed:
$x(t)=v_0*t pm frac{a*t^2}{2} tag{1}$
$v(t)= v_0 pm a*t tag{2}$
kinematics
edited Dec 14 '18 at 16:53
peterh
2,17351731
asked Dec 11 '18 at 20:54
Yusuf123Yusuf123
111
$endgroup$
closed as off-topic by Davide Giraudo, Dando18, John B, KReiser, platty Dec 12 '18 at 0:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, Dando18, John B, KReiser, platty
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
update your question with picture and kinematic equations.
$endgroup$
– yW0K5o
Dec 13 '18 at 21:52
add a comment |
$begingroup$
A car moves along a straight horizontal road from a point A to a point B, where AB = 885 m. The car accelerates from rest at A to a speed of 15 m s −1 at a constant rate a m s−2. The time for which the car accelerates is T seconds. The car maintains the speed of 15 m s−1 for T seconds. The car then decelerates at a constant rate of 2.5 m s−2 stopping at B.
Find the time for which the car decelerates.
How do you approach this question?
Other details:
Here's kinematic equations for distance and speed:
$x(t)=v_0*t pm frac{a*t^2}{2} tag{1}$
$v(t)= v_0 pm a*t tag{2}$
kinematics
edited Dec 14 '18 at 16:53
peterh
2,17351731
asked Dec 11 '18 at 20:54
Yusuf123Yusuf123
111
$endgroup$
A car moves along a straight horizontal road from a point A to a point B, where AB = 885 m. The car accelerates from rest at A to a speed of 15 m s −1 at a constant rate a m s−2. The time for which the car accelerates is T seconds. The car maintains the speed of 15 m s−1 for T seconds. The car then decelerates at a constant rate of 2.5 m s−2 stopping at B.
Find the time for which the car decelerates.
How do you approach this question?
Other details:
Here's kinematic equations for distance and speed:
$x(t)=v_0*t pm frac{a*t^2}{2} tag{1}$
$v(t)= v_0 pm a*t tag{2}$
kinematics
kinematics
edited Dec 14 '18 at 16:53
peterh
2,17351731
asked Dec 11 '18 at 20:54
Yusuf123Yusuf123
111
edited Dec 14 '18 at 16:53
peterh
2,17351731
asked Dec 11 '18 at 20:54
Yusuf123Yusuf123
111
edited Dec 14 '18 at 16:53
peterh
2,17351731
edited Dec 14 '18 at 16:53
peterh
2,17351731
edited Dec 14 '18 at 16:53
peterh
2,17351731
2,17351731
asked Dec 11 '18 at 20:54
Yusuf123Yusuf123
111
asked Dec 11 '18 at 20:54
Yusuf123Yusuf123
111
asked Dec 11 '18 at 20:54
Yusuf123Yusuf123
111
111
closed as off-topic by Davide Giraudo, Dando18, John B, KReiser, platty Dec 12 '18 at 0:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, Dando18, John B, KReiser, platty
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Davide Giraudo, Dando18, John B, KReiser, platty Dec 12 '18 at 0:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, Dando18, John B, KReiser, platty
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
update your question with picture and kinematic equations.
$endgroup$
– yW0K5o
Dec 13 '18 at 21:52
add a comment |
$begingroup$
update your question with picture and kinematic equations.
$endgroup$
– yW0K5o
Dec 13 '18 at 21:52
$begingroup$
update your question with picture and kinematic equations.
$endgroup$
– yW0K5o
Dec 13 '18 at 21:52
$begingroup$
update your question with picture and kinematic equations.
$endgroup$
– yW0K5o
Dec 13 '18 at 21:52
add a comment |
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$begingroup$
update your question with picture and kinematic equations.
$endgroup$
– yW0K5o
Dec 13 '18 at 21:52