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In summary, the conversation discusses how to show that s = ut + .5at^2 given the starting equations of d^2s/dt^2 = a, ds/dt = u, and s = 0 when t = 0. The solution involves deriving v = u + at from the starting equality and incorporating a constant for initial velocity, resulting in the quadratic equation s = u0t + .5at^2.
- #1
Woolyabyss
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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
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- #2
Woolyabyss
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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
- #3
CAF123
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Woolyabyss said:
could you maybe use v = u + at ?
Yes, but first you need to derive this from the starting equality d2s/dt2 = a.
- #4
Woolyabyss
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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.
- #5
rcgldr
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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
- #6
Woolyabyss
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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?
- #7
rcgldr
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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.
- #8
Woolyabyss
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rcgldr said:
The goal here is to produce a quadratic equation, not an intergral with limits.
I got now thanks.
Related to Second Order Differential Equation to show s = ut +(1/2)at^2
1. What is a second order differential equation?
A second order differential equation is a mathematical equation that involves the second derivative of a function. It is commonly used to describe the relationship between a physical quantity and its rate of change over time.
2. How is a second order differential equation used to show s = ut + (1/2)at^2?
In this equation, s represents displacement, u represents initial velocity, a represents acceleration, and t represents time. The second order differential equation is used to show how displacement changes over time due to the initial velocity and acceleration.
3. What is the significance of the (1/2) factor in the equation?
The (1/2) factor represents the constant acceleration of an object. It is multiplied by the time squared to account for the fact that acceleration is constantly changing over time. Without this factor, the equation would only represent a constant velocity.
4. Can this equation be used in any situation involving displacement, velocity, and acceleration?
Yes, this equation can be applied to any situation where an object is experiencing constant acceleration. It is commonly used in physics and engineering to solve problems involving motion and dynamics.
5. Are there any real-world applications of this equation?
Yes, there are many real-world applications of this equation. It can be used to calculate the trajectory of a projectile, the motion of a pendulum, or the acceleration of a falling object. It is also used in the design and analysis of structures, such as bridges and buildings.
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