motion physics (1 Viewer)

jjuunnee · May 17, 2016

s=ut+1/2at^2
v^2=u^2+2as

What are these two formulas used for? I know that the first one can be used to find displacement but it can be figured out through s=vt

Nailgun · May 17, 2016

jjuunnee said:
s=ut+1/2at^2
v^2=u^2+2as

What are these two formulas used for? I know that the first one can be used to find displacement but it can be figured out through s=vt

what if you don't have v?
what if it is accelerating?

Questions typically give you a certain set of variables, or enough info to deduce more variables and you pick an equation to use

the other one is v=u+at

InteGrand · May 17, 2016

jjuunnee said:
s=ut+1/2at^2
v^2=u^2+2as

What are these two formulas used for? I know that the first one can be used to find displacement but it can be figured out through s=vt

$\noindent The first one is helpful to find the (change in) displacement given a time value if we know the object has a constant acceleration and we know its initial velocity $u$. Examples of problems we can solve with this are problems involving for instance how long it'll take a dropped ball to fall a given height under constant gravity of $9.8\text{ m/s}^{2}$.$

$\noindent The second one is also used for constant acceleration and allows us to relate a final velocity to displacement under constant velocity. For example, it'll help if we know we dropped a ball in constant gravity, and we ask how fast it is moving after it has travelled 20 metres.$

$\noindent You should be able to find lots of practice questions online or in a book to help yourself get a feel for how to use these formulas.$

InteGrand · May 17, 2016

jjuunnee said:
s=ut+1/2at^2
v^2=u^2+2as

What are these two formulas used for? I know that the first one can be used to find displacement but it can be figured out through s=vt

$\noindent Also, the first formula doesn't follow \textsl{straight} from $\Delta s = v \Delta t$, since $v$ is varying in when we have a constant acceleration (unless it's a trivial 0 acceleration). The correct formula is: $\Delta s = \bar{v} \Delta t$, where $v$ is the average velocity over a time interval $\Delta t$. This is true regardless of whether the motion has constant acceleration or just general acceleration, because it's just a rearrangement of the definition of average velocity. In the case of constant acceleration though, the $\bar{v}$ can be simplified to: $\bar{v} = \frac{v_i + v_f}{2}$, where $v_i$ is the initial velocity and $v_f$ is the finaly velocity. Hence under constant acceleration $a$, we have$

$$\begin{align*}\Delta s &= \frac{v_i + v_f}{2} \Delta t \\ &= \frac{1}{2}\left(v_i+v_i + a \Delta t\right) \Delta t \quad (\text{using }v_f = v_i + a\Delta t,\text{ from constant acceleration}) \\ &= \frac{1}{2}\left(2v_i + a \Delta t\right) \Delta t \\ &= v_i \Delta t+ \frac{1}{2}a\left(\Delta t \right) ^2 \end{align*}$.$

$\noindent And so we have our formula $s=ut+\frac{1}{2}at^2$, where here $s$ refers to the displacement relative to time 0, and $t$ is the time elapsed ($u$ being initial velocity and $a$ the constant acceleration).$

buckskinbrumby · May 18, 2016

jjuunnee said:
s=ut+1/2at^2
v^2=u^2+2as

What are these two formulas used for? I know that the first one can be used to find displacement but it can be figured out through s=vt

We're using them for projectile motion (space topic, yr12)

motion physics (1 Viewer)

jjuunnee

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Nailgun

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InteGrand

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buckskinbrumby

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