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derivation of kinematic equations
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Question:5] The four kinematic equation has 5 variables: o, , a, t and x. The first does not involve x. The second does not contain ; the third omits t and the last leaves out a. So to complete the set of equations, there should be an equation not involving o. Derive that equation and use the equation to solve the problem below in one step.
The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of 5.60 m/s2 for 4.20 s, making straight skid marks 62.4 m long ending at the tree. With what speed does the car then strikes the tree?
Answers:d = vi * t + .5at^2 vf = vi + at so, rearranging the second equation, vi = at + vf plug that in for vi in the first d=(at + vf)t + .5at^2 From here, just plug&chug and it should result in a final speed of 3.097 m/s towards the tree. Plug it back into the second equation, and you will find that the initial velocity is 26.617 m/s. Not that vi matters, or anything :)
Answers:d = vi * t + .5at^2 vf = vi + at so, rearranging the second equation, vi = at + vf plug that in for vi in the first d=(at + vf)t + .5at^2 From here, just plug&chug and it should result in a final speed of 3.097 m/s towards the tree. Plug it back into the second equation, and you will find that the initial velocity is 26.617 m/s. Not that vi matters, or anything :)
Question:pls show me how to derive the formula of
1.) d=Vit+1/2at^2
using the equations Vf=Vi+at and equation d=1/2 (Vi+Vf)t
2.) V^2=Vi^2+2ad
using the equations d=1/2 (Vi+Vf)t and equation d=Vit+1/2at^2
pls show it step by step so that i can understand clearly..thanks
Answers:Do you know about integrals and primitives? Now seeing the formulas you have there it is very straightforward. d=1/2 (Vi+Vf)t=1/2(vi+vi+at)t=vi.t+1/2at The second formula we derive by eliminating the time t=2d/(vi+vf). d=vi 2d/(vi+vf)+a/2.(2d/(vi+vf)) Simplifying this form you will indeed get: vf =vi +2ad. Yet this yields more of mathematical exercice than real physical reasoning. In fact the first formule would be derived by integrating, as follows: d=o$t o$t a.dt=do+vi.t+at /2 and you have declared do=o.
Answers:Do you know about integrals and primitives? Now seeing the formulas you have there it is very straightforward. d=1/2 (Vi+Vf)t=1/2(vi+vi+at)t=vi.t+1/2at The second formula we derive by eliminating the time t=2d/(vi+vf). d=vi 2d/(vi+vf)+a/2.(2d/(vi+vf)) Simplifying this form you will indeed get: vf =vi +2ad. Yet this yields more of mathematical exercice than real physical reasoning. In fact the first formule would be derived by integrating, as follows: d=o$t o$t a.dt=do+vi.t+at /2 and you have declared do=o.
Question:How would I go about solving the following question?
A flowerpot falls from the ledge of an apartment building. A person in an apartment below, coincidentally in possesion of a highspeed, highprecision timing system, notices that it takes 0.25 s to fall past his 4.5 m tall window. How far above the top of the window is the ledge from which the pot fell?
The answer is 14.4m I am just unsure how to derive this.
Thanks!
Answers:The equation for displacement is x = 4.9*t^2 + Vo*t where t is time and Vo is the initial velocity. The equation for velocity V=a*t where a is acceleration (gravity=9.81m/s^2) Rearrange equation to solve for Vo, giving Vo= (x4.9*t^2)/t Sub in values for x (length of the window), and t. Vo=16.78m/s Use V=a*t to find time taken to travel from rest to Vo, t=16.78/9.81 t=1.71 Now sub that time back into x=4.9*t^2, which gave me a value of 14.3
Answers:The equation for displacement is x = 4.9*t^2 + Vo*t where t is time and Vo is the initial velocity. The equation for velocity V=a*t where a is acceleration (gravity=9.81m/s^2) Rearrange equation to solve for Vo, giving Vo= (x4.9*t^2)/t Sub in values for x (length of the window), and t. Vo=16.78m/s Use V=a*t to find time taken to travel from rest to Vo, t=16.78/9.81 t=1.71 Now sub that time back into x=4.9*t^2, which gave me a value of 14.3
Question:Which kinematics equation, when solved for T, is this:
T = sqrt (2d/A)
Oh yeah and the start velocity is zero.
Can you also tell me, step by step, how to solve for T to get this equation in the first place? Im not that good at algebra 1 and that's what my geometry teacher says too.
Thank You!
Answers:This is the second equation of motion in which you have S, t, a and Vi.. d is distance, t is time, a is acceleration and Vi is initial velocity... d = Vit + 1/2at When you handle problems which have bodies dropped from some height ( free fall ) or which have bodies accelerating from rest then their Initial velocity is zero.. d = (0)t + 1/2at ...............( Any number or variable multiplied by zero = zero ) d = 0 + 1/2at d = 1/2at 2d = at ......(1 is being divided by 2 or right side, so when we move it to left side we multiply it) 2d/a = t ...........................( a is multiplied by t so we write it in division on left side ) Taking sqrt both sides... 2d/a = t .....................( over t is canceled by ) => t = 2d /a
Answers:This is the second equation of motion in which you have S, t, a and Vi.. d is distance, t is time, a is acceleration and Vi is initial velocity... d = Vit + 1/2at When you handle problems which have bodies dropped from some height ( free fall ) or which have bodies accelerating from rest then their Initial velocity is zero.. d = (0)t + 1/2at ...............( Any number or variable multiplied by zero = zero ) d = 0 + 1/2at d = 1/2at 2d = at ......(1 is being divided by 2 or right side, so when we move it to left side we multiply it) 2d/a = t ...........................( a is multiplied by t so we write it in division on left side ) Taking sqrt both sides... 2d/a = t .....................( over t is canceled by ) => t = 2d /a
From Youtube
Kinematics  deriving equations of constant acceleration :A short lesson on basic Kinematics, how to derive the equations of motion under constant acceleration. Check out www.gaussianmath.com for more mechanics and calculus related topics.
Physics: 2D Kinematics: Deriving the Range Equation :This is a basic derivation of the range equation for projectile motion. This equation is useful in a symmetric projectile situation when one wants to find the range when given launch velocity and angle.