Second Order Differential Equation to show s = ut +(1/2)at^2

AI Thread Summary
The discussion focuses on deriving the equation s = ut + (1/2)at^2 from the second order differential equation d^2s/dt^2 = a. Participants discuss integrating the equations, emphasizing the need for an initial velocity constant, u0. The importance of correctly applying limits during integration is highlighted, as it affects the final equation. There is a consensus that the goal is to produce a quadratic equation rather than just an integral with limits. Ultimately, the participants clarify the relationship between velocity, acceleration, and displacement in the context of the problem.
Woolyabyss
Messages
142
Reaction score
1

Homework Statement




If d^2s/dt^2 = a, given that ds/dt = u and s = 0, when t = 0, where a, u are constants

show that s = ut + .5at^2



2. The attempt at a solution

du/dt = a

cross multiplying and then integrating and we get

u = at

ds/dt = at

cross multiply and integrate

s = .5at^2

using limits when t = 0 then s = 0

I can't seem to get out the constant u
 
Physics news on Phys.org
could you maybe use v = u + at ? and say v = u + at but v = ds/dt

so ds/dt = u + at

cross multiplying and integrating and you get

s = ut + .5at^2
 
Woolyabyss said:
could you maybe use v = u + at ?
Yes, but first you need to derive this from the starting equality d2s/dt2 = a.
 
CAF123 said:
Yes, but first you need to derive this from the starting equality d2s/dt2 = a.

I'm not sure how though could I say dv/dt = a

and then I'd get v = at but I just can't seem to get the u.
 
Woolyabyss said:
cross multiplying and then integrating and we get
u = at
After integration there's a constant. Say the initial velocity is u0, then

u = at + u0

The problem statement seems a bit off, if u = ds/dt, then there needs to be a constant for initial velocity in the equation such as u0:

s = u0 t + 1/2 a t^2
 
rcgldr said:
After integration there's a constant. Say the initial velocity is u0, then

u = at + u0

The problem statement seems a bit off, if u = ds/dt, then there needs to be a constant for initial velocity in the equation such as u0:

s = u0 t + 1/2 a t^2

But would using the limits not eliminate the constant of integration?
 
Woolyabyss said:
But would using the limits not eliminate the constant of integration?
The goal here is to produce a quadratic equation, not an intergral with limits.
 
rcgldr said:
The goal here is to produce a quadratic equation, not an intergral with limits.

I got now thanks.
 
Back
Top