Anti-differentiation, can someone explain how they got 80?

Question

luigi0210 · Answer

whpalmer4 · Answer

Quite easy, actually.  Constant deceleration, so $$x=\frac{1}{2}at^2$$$$x=200$$$$a=16$$Solve for $t$.  Then use $$v_f - v_i = at$$with $v_f=0$ and values of $a,t$ already known to find $v_i$

anonymous · Answer

my ans is coming 3200 ft/t^2

parthkohli · Answer

Meh, that may take a long time to type.  I'm cancelling it off.  Do you know how to obtain the following results through calculus?$$v_f = v_i + aT$$$$v_f^2 = v_i^2 + 2ax$$$$x = v_i T + \dfrac{1}{2}aT^2$$This is for constant acceleration.

whpalmer4 · Answer

You can do a quick reasonability check on the answer:

For the first second, the car is going 80 ft/s, decelerating to 80-16=64 ft/s.  Take the midpoint, so the car travels about 72 feet in the first second.  In the second second, it decelerates from 64 to 48 ft/s, midpoint is 56 ft/s, so after 2 seconds, the car has gone about 128 ft.  In the 3rd second, it decelerates from 48 to 32 ft/s, midpoint is 40 ft/s, so after 3 seconds, the car has gone about 168 ft.  In the 4th second, decel. from 32 to 16 ft/s, midpoint is 24 ft/s, so after 4 seconds, total distance is about 192 ft.  Finally, decel from 16 to 0, midpoint is 8 ft/s, after 5 seconds, total distance is 200 ft.  Answer checks out.

whpalmer4 · Answer

Doing it in this fashion works only because we have constant deceleration, of course!

ganeshie8 · Answer

the graph convinces more :)

here is the work :
say, the car took "t" seconds to stop :
$\large  0 - v(0) =   \int \limits_0^t -16 dt$    (acceleration =   velocity')
$\large \implies $
$\large  v(0) = 16t$ ***************


Next, displacement in the interval is 200 :
$\large 200 =  \int \limits_0^t  16t dt$   (velocity =  displacement')
$\large 200 =  16\dfrac{t^2}{2} $
$\large t =  5 $

plug this in $\large v(0)$ :
$\large v(0) = 16t = 16(5) = 80 $

luigi0210 · Answer

Alright, thanks guys! :) 

I haven't seen those formulas before tho, they're not in my either >.<

luigi0210 · Answer

*book either

ganeshie8 · Answer

yeah these are not math related. we just we need to make sense of these somehow :

1)  velocity = rate of change of displacement =  $\large  \dfrac{ds}{dt}$
2)  acceleration= rate of change of velocity =  $\large  \dfrac{dv}{dt}$

ganeshie8 · Answer

working them backwards :

1)  velocity =   $\large  \int a(t)  dt$

2)  position =   $\large    \int v(t)  dt$

whpalmer4 · Answer

The graph shouldn't convince any more than the computation, however — doing the integrals is simply taking the limit of the sum of all of those rectangles, and with a linear function and those rectangles being computed at the midpoint, they are the same.

acceleration is the second derivative of position, so position is the second integral of acceleration.  similarly, velocity is the first derivative of position, so the integral of position is velocity.  I suppose in this context we should really refer to speed rather than velocity, given that we are discussing scalar properties, not vectors.