Do particles have to move in a straight line to apply Suvat equations? I always perceived that they do because acceleration can mean a change in the magnitude of velocity, a change in direction or a change in both. By stating that a particle moves in a straight line with constant acceleration the only "type" of acceleration that can take place is change in the magnitude of velocity.
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2$\begingroup$ never heard of "Suvat", but if the answer is "yes", I recommend not using them. $\endgroup$– JEBCommented Mar 17, 2019 at 22:50
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1$\begingroup$ I think the "kinematic equations" is more commonplace than "suvat equations". I've been studying physics for a significant amount of time and this is the first time I've seen "suvat" being used. $\endgroup$– user113773Commented Mar 17, 2019 at 23:54
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$\begingroup$ "Suvat equations" is a fairly common but minority term. It is handy in-so-far as "kinematic equations" have no a priori reason to be limited to those dealing with constant velocity and constant acceleration, but the suvat equations are defined in this way. None-the-less I don't generally use the term (but I find that being familiar with the usages that go with it indispensable for being able to help students with varied backgrounds). $\endgroup$– dmckee --- ex-moderator kittenCommented Mar 18, 2019 at 1:38
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2 Answers
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Suvat equations come from the consideration that force in the system is constant. In Newtonian mechanics, $$\frac{d\mathbf{v}}{dt}=\mathbf{a} $$ Where, $\mathbf{a}$, is a constant. This gives the result, $$\mathbf{v}=\mathbf{a}t+\mathbf{c} $$ To calculate the integration constant set $t=0$, and you shall have $\mathbf{c}=\mathbf{v}_0$. The equation becomes, $$\mathbf{v}=\mathbf{a}t+\mathbf{v}_0 $$ Similarly we know that, $$\mathbf{v}=\frac{d\mathbf{r}}{dt}=\mathbf{a}t+\mathbf{v}_0 $$ Which gives us, $$\mathbf{r}=\frac{1}{2}\mathbf{a}t^2+\mathbf{v}_0t+\mathbf{c} $$ To get the integration constant, set $t=0$ and the constant will be $\mathbf{c}=\mathbf{r}_0 $ which will give us, $$\mathbf{r}=\frac{1}{2}\mathbf{a}t^2+\mathbf{v}_0t+\mathbf{r}_0 $$ You can get the third equation by taking dot product of $\mathbf{v}$ and doing some intelligent substitution to remove time variable. Any other variation can be achieved in similar ways. Also this makes your assumption kind of right, this equation holds for particle moving in a straight line, since by moving in straight line they are exhibiting constant acceleration. However, there can also be cases when acceleration is constant, and yet motion is not in a straight line. Obviously situations where the acceleration changes direction with the object, like a car turning around, will then move outside the domain of suvat equation since there is a centripetal force working on it which is not constant in direction. However when considered the force lies outside the object, for cases like projectile motion under constant gravity condition, the object's path will not be straight, however you can apply suvat equations. Although for any sort of constant circular acceleration, that is constant angular acceleration, like the discussed car's case, or spinning pebble attached with the string, one can come up with equations that corresponds to suvat's case in constant acceleration.
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$\begingroup$ I think it should be made clear that one can apply them to projectile motions only after decomposing them into components along 2 straight, perpendicular directions? $\endgroup$ Commented Mar 24, 2022 at 14:04
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The short answer is no. What you call "suvat" equations is the consequence of a constant aceleration (magnitude). This is different to the assumption of a constant force. For instance consider a charged particle moving under the influence of a magnetic field, the force would not be constant (it changes direction while the magnitude remains unchanged).
A particle as described above would move in a circular path, and the "suvat"equations would still hold (eventhough they would be now called the "$\theta \omega_0 \omega_f \alpha t$" equations.
