How Can You Derive the Kinematic Equation V2x^2 = V1x^2 + 2ax(delta)x?

AI Thread Summary
To derive the kinematic equation V2x^2 = V1x^2 + 2ax(delta)x, start with the equations V2x = V1x + ax(delta)t and (delta)x = V1x(delta)t + 1/2ax(delta)t^2. Solve the first equation for time, then substitute this expression into the second equation to eliminate time. Upon simplification, the delta t term will be removed, leading to the desired equation. It's noted that using simplified notation without the x's can make the derivation clearer, as these are standard one-dimensional motion formulas.
Intagral
Messages
1
Reaction score
0
The Problem:
Using V2x=V1x+ax(deltat)t and (delta)x=V1x(delta)t + 1/2ax(delta)t^2 derive the formula V2x^2=V1x^2+2ax(delta)t


Homework Equations


1st and 3rd x-component Kinematics Equations


The Attempt at a Solution


Ok, so I attempted to solve it, but it didn't work. I don't want the answer but could someone just push me in the right direction.
 
Physics news on Phys.org
Solve the v2x = equation for time. Substitute that expression into the other equation to eliminate time. Simplify. You will find that
V2x^2=V1x^2+2ax(delta)t
will not have the delta t at the end.

By the way, much easier if you leave out all those x's. These are just the one dimensional accelerated motion formulas from grade 11:
V = Vi + at and d = Vi*t + 0.5*a*t^2
and the one you are looking for is V^2 - Vi^2 = 2*a*d
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top