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How do you calculate an equation that asks to calculate the final velocity without time being given? But with distance being given...

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    $\begingroup$ You've got 4 kinematic equations, take a look at them. $\endgroup$ – Kyle Kanos Oct 13 '14 at 19:21
  • $\begingroup$ Im in grade 9... So I don't actually understand could you explain that in the simplest way possible? $\endgroup$ – lil'kay Oct 13 '14 at 19:29
  • $\begingroup$ You should have been given, by your teacher, a set of 4 equations involving distance ($d$), velocity ($v$), acceleration ($a$) and time ($t$). One of these 4 equations does not have $t$ in it. $\endgroup$ – Kyle Kanos Oct 13 '14 at 19:35
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    $\begingroup$ $v^2 = u^2 + 2as$ possibly? $\endgroup$ – John Rennie Oct 13 '14 at 19:35
  • $\begingroup$ Or possibly five equations.. en.wikipedia.org/wiki/… $\endgroup$ – DJohnM Oct 14 '14 at 1:13
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Assuming you are considering motion with constant acceleration $a$ and initial velocity $v_0$, you know: $$s=v_0 t+\frac{1}{2}a t^2$$ You can solve that quadratic equation for t: $$t=-\frac{v_0}{a}+\sqrt{\frac{v_0^2}{a^2}+\frac{2s}{a}}$$ Your final velocity is then $$v=v_0+at=\sqrt{v_0^2+2as}$$. In the case that the motion started from rest: $$v=\sqrt{2as}$$ Keep in mind that all this is only correct for constant acceleration.

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There are two more kinematics equations: 1. Non accelerated motion – that is motion at a constant velocity

The area under the line of the velocity–time graph is the distance travelled by the object in the time t.

For example u = 20m/s and t = 300 s Distance (s) = ut = 20 x 300 = 6000 m

The equation for non accelerated motion is:

Distance (s) = velocity (u or v) x time (t) s = vt

  1. Accelerated motion Acceleration (a) = [change in velocity]/time = [v - u]/t or a = [v - u]/t

Another version is:

v = u + at

Distance travelled = area under the line = ut + ½ (v-u)t

But acceleration = (v-u)/t and so (v-u) = at therefore:

Distance travelled (s) = ut + ½ (v-u)t = ut + ½ [at]t = ut + ½ at2

s = ut + ½ at2

If the object starts from rest u = 0 and so the equation becomes:

s = ½ at2

Another useful equation is:

v2 = u2 + 2as

This equation can be proved as follows: v = u + at therefore t = (v-u)/a but s = ut + ½ at2 and so

s = ut + ½ a([v-u]/a)2 therefore: 2s = 2u(v-u)/a + (v2 – 2uv + u2)/a

So: 2as = 2uv – 2u2 + v2 – 2uv + u2 and so v2 = u2 + 2as

Equations summary

CONSTANT VELOCITY 1. s = vt

CONSTANT ACCELERATION 2. v = u + at 3. s = ut + ½ at2 4. average velocity = [v + u]/t 5. v2 = u2 + 2as

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