Constant Acceleration formulae 2
SUVAT Equations
- \(s\) is displacement
- \(u\) is initial velocity
- \(v\) is final velocity
- \(a\) is acceleration
- \(t\) is time
Equations
\[s=t(\frac{v+u}{2})$$
$$v=u+at$$
$$v^2=u^2+2as$$
$$s=ut+\frac{1}{2}at^2$$
$$s=vt-\frac{1}{2}at^2\]
Example 8
A particle is moving in a straight horizontal line with constant deceleration 4 m s-2. At time t = 0 the particle passes through a point O with speed 13 m/s travelling towards a point A, where OA = 20m. Find:
a) the times when the particle passes through A
b) the value of t when the particle returns to O.
a)
- \(u=13\)
- \(s=20\)
- \(a=-4\)
- \(t=?\)
- \(s=ut+\frac{1}{2}at^2\)
- \(20=13t-\frac{1}{2}\times 4t^2\)
- \(= 13t-2t^2\)
- \(2t^2-13t+20=0\)
- \(t=\frac{5}{2}\) or \(t=4\)
b)
- Finding when \(s=0\)
- \(s=0\)
- \(u=13\)
- \(a=-4\)
- \(t=?\)
- \(s=ut+\frac{1}{2}at^2\)
- \(0=13t-2t^2\)
- \(0=t(13-2t)\)
- So \(t=0\) and \(t=\frac{13}{2}\)
- So the particle passes back through \(O\) after \(6.5\) seconds.