Deriving the Equation v^2=U^2+2as using Physics Principles

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In summary, by substituting t=(v-u)/a into the equation s=ut+1/2at^2, we obtain s=u((v-u)/a)+1/2a((v-u)/a)^2. By simplifying this equation, we can show that v^2=U^2+2as.
  • #1
stuwalshe
Hi, I need help with the following, I don't think it is too hard but I don't know how to approach it.

given that v=u+at and s=ut+1/2at^2

show that v^2=U^2+2as

thankyou
 
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  • #2
Have you tried anything yet? Please note that for homework questions you must show your work before we can help you.

Here's a hint: look at the variables in the equation you are trying to show. There is one variable that is present in the first two equations, but not in the final one. Can you write one of the first two in a way such that you can eliminate this variable?
 
  • #3
cristo
I can see that t is common to both the first equations but not in the 3rd, I guess they cancel out some how, I have made t the subject of the first equation. i.e

v=u+at == t=(v-u)/a.

I think I then have to substitute this into the second equation, this is where I come unstuck, I am unsure of how to substitute the 2 equations together.
t=(v-u)/a
s=ut+1/2at^2

do I then multiply both sides of each equation? i.e

st=ut+1/2at^2(v-u)/a ?

if so then I am unsure how to eliminate the t's and how to get the u^2
thankyou
 
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  • #4
stuwalshe said:
cristo
I can see that t is common to both the first equations but not in the 3rd, I guess they cancel out some how, I have made t the subject of the first equation. i.e

v=u+at == t=(v-u)/a.
Good.

I think I then have to substitute this into the second equation, this is where I come unstuck
Good idea. The easiest way to proceed is to take t=(v-u)/a and substitute it directly into the equation s=ut+1/2at^2. So, you would obtain [tex]s=u\cdot\left(\frac{(v-u)}{a}\right)+\frac{1}{2}a\left(\frac{(v-u)}{a}\right)^2[/tex]

Can you continue from here by simplifying?
 
  • #5
thanks for your help, could you please take me back a step and explain how you got rid of the t's,

For some reason I am really struggling with this one, when I look to simplify the expression you came up with, I can't seam to think of how to get to where I want to be.
do you think you could start me off.

stu
 
  • #6
sorry mate, I was being silly, I understand how the t's cancel, like you said you just substituted them, I think I have done too much today, I still can't see how to simplify the equation though, if you could start me off that would be great.

stu
 
  • #7
Well, I presume you can multiply out the first term. For the second term take out a^2 from the denominator, and then expand (v-u)^2 using whichever technique you have learnt.
 
  • #8
i expanded to

2as=(uv-u^2)/au + (v^2-2uv+v^2)/a^2

I do not think this is correct though because nothing seams to cancel
sorry about this
stu
 
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FAQ: Deriving the Equation v^2=U^2+2as using Physics Principles

1. What is the significance of the equation v^2=U^2+2as in physics?

The equation v^2=U^2+2as is known as the second equation of motion and is a fundamental equation in classical mechanics. It describes the relationship between an object's initial velocity (U), final velocity (v), acceleration (a), and displacement (s).

2. How is the equation v^2=U^2+2as derived?

The equation v^2=U^2+2as is derived from the equations of motion using the principles of calculus. It is a result of integrating the equation for acceleration, a=dv/dt, twice with respect to time.

3. What are the units of measurement for each variable in the equation v^2=U^2+2as?

The units for velocity (v) and initial velocity (U) are typically meters per second (m/s), while acceleration (a) is measured in meters per second squared (m/s^2). Displacement (s) is measured in meters (m).

4. How is the equation v^2=U^2+2as used in real-life applications?

The equation v^2=U^2+2as is used to predict the motion of objects under constant acceleration. It has many practical applications in fields such as engineering, sports, and transportation. For example, it can be used to calculate the trajectory of a projectile or the braking distance of a car.

5. Are there any limitations to the equation v^2=U^2+2as?

Yes, the equation v^2=U^2+2as assumes that the acceleration remains constant throughout the motion of an object. This may not always be the case in real-life situations, such as when an object experiences varying levels of friction or air resistance. Additionally, the equation is only applicable in the absence of external forces, such as gravity.

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