How were these linear motion equations derived?

In summary, two equations were derived in this conversation. The first equation is derived by integrating the equation for constant acceleration twice. It is represented by x = xo + vot + 1/2at2 and explains the relationship between position, initial velocity, acceleration, and time. The second equation, v2 = vo2 + 2a(xf-xi), is derived from the first equation and further explains the relationship between velocity, initial velocity, acceleration, and position. The 1/2 in the first equation comes from the constant acceleration being divided by 2. However, it should be noted that the formula for the second equation may only hold if the initial velocity is equal to the velocity at time 0, so further clarification
  • #1
jshpark
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how were these equations derived or found?

x = xo + vot + 1/2at2I tried to use v=(xf-xi)/(tf-ti) and a=(vf-vi)/(tf-ti)

but I don't know where the 1/2 comes from

also this one:
v2 = vo2 + 2a(xf-xi)
 
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  • #2
The first one is what you get if you integrate the equation that describes constant acceleration twice. If the acceleration is constant (=a), we have
$$x''(t)=a$$ This implies that there's a constant ##v_0## such that $$x'(t)=at+v_0.$$ The constant is denoted by ##v_0## because ##x'(t)## is the velocity at time t, and the result we just found implies that ##x'(0)=v_0##, so ##v_0## is the velocity at time 0. The result we found also implies that there's a constant ##x_0## such that
$$x(t)=\frac 1 2 at^2+v_0t+x_0.$$ The constant is denoted by ##x_0## because it's equal to ##x(0)##, the position at time 0.

The other one follows from the result for the velocity above, and
$$v(t)-v_0=\int_0^t x''(t)dt=at.$$ These results imply that
$$v(t)^2=(v_0+at)^2=v_0{}^2+a^2t^2+2v_0at=v_0{}^2+2a\bigg(\frac 1 2 at^2+v_0t\bigg)=(v_0)^2+2a(x(t)-x_0).$$

I think the formula you wrote down only holds if ##v(t_i)=v_0##, so you may want to check your book to see if it says exactly what you wrote, or something slightly different.
 
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FAQ: How were these linear motion equations derived?

What are the basic equations for linear motion?

The basic equations for linear motion are:

  • Velocity (v) = Distance (d) / Time (t)
  • Acceleration (a) = Change in Velocity (Δv) / Time (t)
  • Displacement (Δd) = Initial Velocity (v0) x Time (t) + 1/2 x Acceleration (a) x Time (t)2
  • Final Velocity (v) = Initial Velocity (v0) + Acceleration (a) x Time (t)
  • Distance (d) = Initial Velocity (v0) x Time (t) + 1/2 x Acceleration (a) x Time (t)2

What is the difference between speed and velocity in linear motion?

Speed is a scalar quantity that refers to how fast an object is moving, while velocity is a vector quantity that refers to both the speed and direction of an object's motion.

How do I calculate the acceleration of an object in linear motion?

To calculate acceleration, you can use the equation: Acceleration (a) = Change in Velocity (Δv) / Time (t). You will need to measure the change in velocity over a specific time period to determine the acceleration.

Can linear motion equations be used for both constant and changing acceleration?

Yes, linear motion equations can be used for both constant and changing acceleration. However, for changing acceleration, you will need to use calculus to determine the acceleration at different points in time.

How can I use linear motion equations to predict the future position of an object?

To predict the future position of an object using linear motion equations, you will need to know the initial velocity, acceleration, and time. You can then use the equation: Displacement (Δd) = Initial Velocity (v0) x Time (t) + 1/2 x Acceleration (a) x Time (t)2 to calculate the object's future position. However, this assumes that the object's acceleration remains constant throughout the time period.

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