Mixing
To calculate the final velocity, (v) of a moving body that had an initial velocity, (u)…
$begingroup$
...and had been under constant acceleration, $a$ for a period of time, $t$ the following formula is used:
$$v = u + at$$
so the following terms are constant,initial vel.$u$ and the constant acceleration, $a$.
Given the definition of velocity is the rate of change distance in a constant direction and assuming the body is moving in a straight line.
i.e. $$v = frac{ds}{dt}$$ where $s$ is distance travelled
Integrate $v$ with respect to time $t$ to find the formula for calculating the distance $s$ travelled in a time $t$
integration
edited Jan 29 at 18:14
SNEHIL SANYAL
656110
asked Jan 29 at 16:16
Rigers MehmetajRigers Mehmetaj
12
$endgroup$
add a comment |
$begingroup$
...and had been under constant acceleration, $a$ for a period of time, $t$ the following formula is used:
$$v = u + at$$
so the following terms are constant,initial vel.$u$ and the constant acceleration, $a$.
Given the definition of velocity is the rate of change distance in a constant direction and assuming the body is moving in a straight line.
i.e. $$v = frac{ds}{dt}$$ where $s$ is distance travelled
Integrate $v$ with respect to time $t$ to find the formula for calculating the distance $s$ travelled in a time $t$
integration
edited Jan 29 at 18:14
SNEHIL SANYAL
656110
asked Jan 29 at 16:16
Rigers MehmetajRigers Mehmetaj
12
$endgroup$
$begingroup$
What exactly are you trying to ask here?
$endgroup$
– Hyperion
Jan 29 at 16:18
$begingroup$
You just have to integrate the function $v(t) = u + a t$ if that's what you're asking.
$endgroup$
– Harnak
Jan 29 at 16:21
add a comment |
$begingroup$
...and had been under constant acceleration, $a$ for a period of time, $t$ the following formula is used:
$$v = u + at$$
so the following terms are constant,initial vel.$u$ and the constant acceleration, $a$.
Given the definition of velocity is the rate of change distance in a constant direction and assuming the body is moving in a straight line.
i.e. $$v = frac{ds}{dt}$$ where $s$ is distance travelled
Integrate $v$ with respect to time $t$ to find the formula for calculating the distance $s$ travelled in a time $t$
integration
edited Jan 29 at 18:14
SNEHIL SANYAL
656110
asked Jan 29 at 16:16
Rigers MehmetajRigers Mehmetaj
12
$endgroup$
...and had been under constant acceleration, $a$ for a period of time, $t$ the following formula is used:
$$v = u + at$$
so the following terms are constant,initial vel.$u$ and the constant acceleration, $a$.
Given the definition of velocity is the rate of change distance in a constant direction and assuming the body is moving in a straight line.
i.e. $$v = frac{ds}{dt}$$ where $s$ is distance travelled
Integrate $v$ with respect to time $t$ to find the formula for calculating the distance $s$ travelled in a time $t$
integration
integration
edited Jan 29 at 18:14
SNEHIL SANYAL
656110
asked Jan 29 at 16:16
Rigers MehmetajRigers Mehmetaj
12
edited Jan 29 at 18:14
SNEHIL SANYAL
656110
asked Jan 29 at 16:16
Rigers MehmetajRigers Mehmetaj
12
edited Jan 29 at 18:14
SNEHIL SANYAL
656110
edited Jan 29 at 18:14
SNEHIL SANYAL
656110
edited Jan 29 at 18:14
SNEHIL SANYAL
656110
656110
asked Jan 29 at 16:16
Rigers MehmetajRigers Mehmetaj
12
asked Jan 29 at 16:16
Rigers MehmetajRigers Mehmetaj
12
asked Jan 29 at 16:16
Rigers MehmetajRigers Mehmetaj
12
12
$begingroup$
What exactly are you trying to ask here?
$endgroup$
– Hyperion
Jan 29 at 16:18
$begingroup$
You just have to integrate the function $v(t) = u + a t$ if that's what you're asking.
$endgroup$
– Harnak
Jan 29 at 16:21
add a comment |
$begingroup$
What exactly are you trying to ask here?
$endgroup$
– Hyperion
Jan 29 at 16:18
$begingroup$
You just have to integrate the function $v(t) = u + a t$ if that's what you're asking.
$endgroup$
– Harnak
Jan 29 at 16:21
$begingroup$
What exactly are you trying to ask here?
$endgroup$
– Hyperion
Jan 29 at 16:18
$begingroup$
What exactly are you trying to ask here?
$endgroup$
– Hyperion
Jan 29 at 16:18
$begingroup$
You just have to integrate the function $v(t) = u + a t$ if that's what you're asking.
$endgroup$
– Harnak
Jan 29 at 16:21
$begingroup$
You just have to integrate the function $v(t) = u + a t$ if that's what you're asking.
$endgroup$
– Harnak
Jan 29 at 16:21
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
$$s=int v dt=int(u+at) dt=ut+frac12at^2$$
Recognise this?
answered Jan 29 at 16:26
Rhys HughesRhys Hughes
7,0801630
$endgroup$
$begingroup$
Thank your got it!
$endgroup$
– Rigers Mehmetaj
Jan 29 at 16:40
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
$begingroup$
$$s=int v dt=int(u+at) dt=ut+frac12at^2$$
Recognise this?
answered Jan 29 at 16:26
Rhys HughesRhys Hughes
7,0801630
$endgroup$
$begingroup$
Thank your got it!
$endgroup$
– Rigers Mehmetaj
Jan 29 at 16:40
add a comment |
$begingroup$
$$s=int v dt=int(u+at) dt=ut+frac12at^2$$
Recognise this?
answered Jan 29 at 16:26
Rhys HughesRhys Hughes
7,0801630
$endgroup$
$begingroup$
Thank your got it!
$endgroup$
– Rigers Mehmetaj
Jan 29 at 16:40
add a comment |
$begingroup$
$$s=int v dt=int(u+at) dt=ut+frac12at^2$$
Recognise this?
answered Jan 29 at 16:26
Rhys HughesRhys Hughes
7,0801630
$endgroup$
$$s=int v dt=int(u+at) dt=ut+frac12at^2$$
Recognise this?
answered Jan 29 at 16:26
Rhys HughesRhys Hughes
7,0801630
answered Jan 29 at 16:26
Rhys HughesRhys Hughes
7,0801630
answered Jan 29 at 16:26
Rhys HughesRhys Hughes
7,0801630
answered Jan 29 at 16:26
Rhys HughesRhys Hughes
7,0801630
7,0801630
$begingroup$
Thank your got it!
$endgroup$
– Rigers Mehmetaj
Jan 29 at 16:40
add a comment |
$begingroup$
Thank your got it!
$endgroup$
– Rigers Mehmetaj
Jan 29 at 16:40
$begingroup$
Thank your got it!
$endgroup$
– Rigers Mehmetaj
Jan 29 at 16:40
$begingroup$
Thank your got it!
$endgroup$
– Rigers Mehmetaj
Jan 29 at 16:40
add a comment |
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$begingroup$
What exactly are you trying to ask here?
$endgroup$
– Hyperion
Jan 29 at 16:18
$begingroup$
You just have to integrate the function $v(t) = u + a t$ if that's what you're asking.
$endgroup$
– Harnak
Jan 29 at 16:21