Yahoo Answers: Answers and Comments for How to solve for time given distance and acceleration using calculus? [Mathematics]
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From angel
enUS
Wed, 24 Jun 2009 11:52:59 +0000
3
Yahoo Answers: Answers and Comments for How to solve for time given distance and acceleration using calculus? [Mathematics]
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From mandynicole_03: Well, the position equation for this is a*t^2 ...
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Wed, 24 Jun 2009 12:02:31 +0000
Well, the position equation for this is a*t^2 + v*t  h0, where h0 is the initial height, a is acceleration, and v is initial velocity. So, this position equation is 32.2*t^2  100. You set this equal to zero because you want the time that the height (or position) is 0 ft off the ground. So...
0 = 32.2*t^2  100
100 = 32.2*t^2 Divide by 32.2, take the (positive) square root, and you have the time when it hits the ground

From grunfeld: Hint: if you drop a ball regardless of height...
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Wed, 24 Jun 2009 11:59:50 +0000
Hint: if you drop a ball regardless of height, what is your initial velocity?
s = 16.1t^2 + 100
 100 = 16.1t^2
100 / 16.1 = t^2
10 / sqrt(16.1) = t

From Caps Lock (TRS_BC): Use the equation:
distance = a*t^2 / 2 = g*...
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Wed, 24 Jun 2009 11:58:38 +0000
Use the equation:
distance = a*t^2 / 2 = g*t^2 / 2
t = √(2*distance/g)

From Sara: Hi he.goes.the.distance Use the formula v^u^2...
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Mon, 11 Apr 2016 06:24:41 +0000
Hi he.goes.the.distance Use the formula v^u^2 = 2as (=2as when it is acceleration) v= 0 a = 16 s = 200 o^2  u^2 = 2*16*200 u = 80 ft/sec Shy

From vij: D= Vi(t) + 1/2 a(t^2) In your question D = 1...
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Wed, 24 Jun 2009 12:07:56 +0000
D= Vi(t) + 1/2 a(t^2) In your question D = 100 ft, Vi = 0,
a = 32.2 ft/ sec^2
Sub the values in the equation and solve for t.
100 = 0(t) + 1/2(32.2) t^2
100 = 16.1 x t^2
100/ 16.1 = t^2
6.2 = t^2
2.5 = t
So ball takes 2.5 sec to hit the ground.

From Anonymous: Since this is a Calculus problem, we will star...
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Wed, 24 Jun 2009 12:03:06 +0000
Since this is a Calculus problem, we will start with
dV/dt = a
where
dV/dt = acceleration of the ball as it is dropped from top of the building
a = acceleration due to gravity = 32.2 ft/sec^2 (constant)
The above equation then becomes
dV/dt = 32.2
V = ⌠32.2(dt)
and integrating the above,
V = 32.2t + C
where
C = integration constant
Since the ball was dropped at t = 0, then
0 = 0 + C and C = 0, hence the equation for V becomes
V = 32.2t
Since V = dS/dt, then
dS/dt = 32.2t
and
S = ⌠(32.2t)dt
Integrating the above,
S = (1/2)(32.2t^2) + C
and when t = 0, the ball has not moved yet, so
0 = 0 + C
and the above equation for "S" simplifies to
S = 16.1t^2
When S = 100 (distance travelled by the ball),
100 = 16.1t^2
and solving for "t"
t = 2.49 sec
Hope this helps.

From hsueh010: a = dv/dt = 32.2 ft/s²
so integrate
v = 32...
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Wed, 24 Jun 2009 11:59:50 +0000
a = dv/dt = 32.2 ft/s²
so integrate
v = 32.2ft/s² t + v(0)
but you said v(0) = 0 so
v = (32.2ft/s²) t
but v = ds/dt
so
ds/dt = (32.2ft/s²) t
s = (32.2ft/s²)(1/2) t² + s(0)
so you want to know when s = 0 (final value on ground) when s(0) (initial position)= 100ft
0 = (16.1ft/s²) t² + 100ft
Solve for t!
[100/16.1]s² = t²
6.2 s² = t²
t = 2.489 sec