# Why do two formulas $d = v_0 t + \frac{at^2}{2}$ and $d=vt$ yield different results?

All of you are sure to be familiar with these two equation:

$d = v_0 t + \frac{at^2}{2}$ and $d=\overline vt$

Given the same initial and final velocities, and time and acceleration. With the second equation I need not use the acceleration. But the distance found using these equations are different!

For this who say that the second equation assume a = 0, it isn't always true. I use the formula for graphs of uniformly accelerated motion to find the distance travelled at the end of certain time. I use initial and final velocities to find the average velocity and multiply the velocity by the time in second on the graph.

Wait, I just tested another uniformly accelerated motion problem with both formulas. The result are the same! What? This is only for some specific problems.

• One equation assumes a non-zero acceleration, the other does not. In fact, the second equation should be $\overline v$ because it is an average velocity, not an instantaneous one
– Sean
Oct 27, 2014 at 0:27
• @Sean: Or one can just say that the second equation holds only for constant-velocity motion.
– user4552
Oct 27, 2014 at 1:46
• I almost did, but I didn't in the interest of clarity because that's not completely true true. You can use $\overline v= \frac{\Delta x}{\Delta t}$ any time you know the average velocity over the time interval, even if the instantaneous velocities over the time interval are not the same, i.e. constant.
– Sean
Oct 27, 2014 at 1:58
• @Sean: OK, formulating the logic more carefully: if $v$ is constant, then $d=vt$ is true. This if-then is a true statement and avoids introducing the notion of average velocity, which is essentially never of any interest in physics.
– user4552
Oct 27, 2014 at 2:26
• @Sean If there is constant acceleration, then the average velocity cannot be found using $\bar v=\frac {\Delta x}{\Delta t}$.
– LDC3
Oct 27, 2014 at 3:24

If you have a uniform acceleration the average velocity $\bar{v}$ is just $$\bar{v}=v_0+\frac{at}{2}.$$ because the final velocity is $v_{initial}=v_0$ and $v_{final}=v_0+at$, i.e. $$\bar{v}=\frac{1}{2}(v_{initial}+v_{final})=v_0+\frac{at}{2}.$$ Then you get $$d=\bar{v}t=v_0 t+\frac{at^2}{2}.$$

The second formula is a special case of the first one where the acceleration is zero. If you substitute $a=0$ into the first formula you get the second one, as is expected

Don't get too stuck on formulas.

More generally: (1) velocity is the time rate of change of position and (2) acceleration is the time rate of change of velocity

So given any (1 dimensional) function of x(t), v(t) or a(t) to begin with, the calculus constrains all remaining relationships.

The formulas you wrote are for special situations: the first assumes constant acceleration and an initial velocity. The second position with presumably variable velocity.

• I don't really understand what you said about the calculus Oct 27, 2014 at 19:48
• integral/differential calculus is a system of mathematics that deals with infinitesimal differences and sums. Developed by Newton and Leibniz in the 17th century it offered scientists a way of modeling kinematic and dynamic systems. Position, velocity and acceleration are related by differential or integral calculus. Calculus is used to derive the algebraic equations you see in HS and College textbooks. Some HS courses work within calculus, others only algebra. Oct 27, 2014 at 20:24
• What I actually meant to say: Integral/differential calculus is a system of mathematics that deals with infinitesimal differences and sums. Developed by Newton and Leibniz in the 17th century it offered scientists a way of modeling kinematic and dynamic systems. Position, velocity and acceleration are related by differential or integral calculus. Calculus is used to derive the algebraic equations you see in HS and College textbooks. Some HS courses in physics work directly with calculus, others only with the algebraic formulas. Oct 27, 2014 at 20:34

Actually, a more correct full formula would be: $$x = x_0 + v_0t + \frac{1}{2}at^2$$ Where I have used $x$ instead of $d$. Note that $x_0$ and $v_0$ are fixed constant values.

• Can u explain what x sub 0 is? Oct 27, 2014 at 2:46
• Why is it?????? Oct 27, 2014 at 2:47
• The formula yields the resultant position x(t) given an initial position x sub 0 and an initial velocity v sub 0 and the acceleration a. Oct 27, 2014 at 2:49

D=vt is used when acceleration is zero but D=ut+1/2 at^2 is used when there is a constant acceleration.