vf=vi+at equation can be solved by substituting the letters in the equation with there actual values where vf is the finall velocity, vi is the initial velocity, a is the acceleration and t is the time.
How you solve v(f) = v(i) + at (or even slove it) depends on which of the variables are known and which one you need to solve for. The steps will differ accordingly.
f=ma vf=vi+at s=vi+1/2at
A equals Vf minus Vi divided by time equals triangle v divided by time
vf2 = vi2 + 2ad, where vf is final velocity, vi is initial velocity, a is acceleration, and d is displacement. Solve for a.vf = vi + at, where t is time time. Solve for a.
(vf-vi)/ t is ?
The ball reaches a maximum height of ~81.6 meters. How to do it: Find the change in time using Vf = Vi -g(t) We know: Vf = 0 (terminal velocity) and Vi = 40 (given in question) So, Vf = Vi - g(t) => 0 = 40 - g(t) --Rework the equation to solve for t... t = 40/g => since g = 9.8... t = 40/9.8 ~ 4.08 So the change in time is about 4.08 seconds Now plug your new values into the kinematics equation Yf = Yi + Vi(t) + 1/2(a)(t^2) Plug in our values and solve => Yf = 0 + (40 * 4.08) - (9.8 * 4.08^2)/2 Therefore, Yf = 81.6 meters. This is your maximum height. Remember to find the time taken for the ball to fall from this height back to the ground if you need to calculate the total time of the fall. Source: I recently took a test with this exact question, got it right. Thanks for posting! ~Secret Physics Man
Yes recalling the first equation of motion ie Vf = Vi + at Here Vf is final velocity and Vi is the initial velocity. a the acceleration and t is the time Now taking at on the other side ie left side we get Vf - at = Vi This is what mentioned here.
f=ma vf=vi+at s=vi+1/2at
Vf = Vi + at Where Vf = final velocity Vi = initial velocity a = acceleration t = time
1) Work input = Force * distance 2) Force = mass*acceleration 3) Acceleration = (Vf - Vi) ÷ time 4) Force = mass * [(Vf - Vi) ÷ time] 5) Distance = Average velocity * time 6) Average velocity = (Vf + Vi) ÷ 2 7) Distance = [(Vf + Vi) ÷ 2] * time Eq#4 * EQ #7 8) Work input = mass * [(Vf - Vi) ÷ time] * [(Vf + Vi) ÷ 2] * time Time cancels 9) Work input = mass * (Vf - Vi) * (Vf + Vi) ÷ 2 10)(Vf - Vi) * (Vf + Vi) = Vf^2 - Vi^2 11)Work input = mass * [Vf^2 - Vi^2] ÷ 2 12)Work input = mass *( Vf^2 ÷ 2) - mass * (Vi^2 ÷ 2) 13)Kinetic energy = ½ mass *velocity ^2 14) Change in KE = (½ mass * Vf ^2) - (½ mass * Vi ^2) Equation #12 = Equation #14 so 15)Work input = Change in KE 16)Work input = ∆ KE
when using energy use the kinetic energy equation for change KE = .5(M)(Vf^2 - Vo^2) M = mass Vf = fianal velocity Vo = initial velocity
A equals Vf minus Vi divided by time equals triangle v divided by time
Well this could be a one step or 2 step question D= (vi + vf)/2 x t an solve for time or...2 step v^2= vi^2 + 2ad: once you have found a, use: V=vi + at then solve for time
The VF in VF Corp. stands for Vanity Fair.
W=Fd and F=ma W is work (energy), F is force, d is distance, m is mass, and a is acceleration. plug in to get W=mad so ad=W/m (vf^2)=(vi^2) + 2ad where vf and vi are velocity final and initial respectively. Assume vi=o so (vf^s)=2ad rearrange so ad=(vf^2)/2 Plug in again to get W=m[(vf^2)/2] which is kinetic energy.
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VF=version francaise= voila
Two ways: If the change in velocity is the result of hitting something, use the Momentum Equation. If the change in velocity is the result of applying a force, use the Impulse Equation. You probably mean this equation, which is: FT = m(Vf - Vo) Or, An object of mass "m" will change from velocity "Vo" to velocity "Vf" if the force "F" is applied for "T" seconds.