A brief note on Equations of Motion . Read it before a test. Try each question at the end before you open its answer.
The big ideas For steady acceleration : v = u + a t v = u + at v = u + a t , s = u t + 1 2 a t 2 s = ut + \tfrac12 at^2 s = u t + 2 1 a t 2 , v 2 = u 2 + 2 a s v^2 = u^2 + 2as v 2 = u 2 + 2 a s . Choose the equation with your three known quantities and the one you want. Free fall: without air, everything falls with g = 9.8 m/s 2 g = 9.8\ \text{m/s}^2 g = 9.8 m/s 2 . Thrown up at u u u : time to the top u g \dfrac{u}{g} g u , greatest height u 2 2 g \dfrac{u^2}{2g} 2 g u 2 ; coming down takes as long. Circular motion at a steady speed is accelerated: v = 2 π r T v = \dfrac{2\pi r}{T} v = T 2 π r , velocity along the tangent. The facts at a glance You do not know Use s s s v = u + a t v = u + at v = u + a t v v v s = u t + 1 2 a t 2 s = ut + \tfrac12 at^2 s = u t + 2 1 a t 2 t t t v 2 = u 2 + 2 a s v^2 = u^2 + 2as v 2 = u 2 + 2 a s a a a s = 1 2 ( u + v ) t s = \tfrac12(u + v)t s = 2 1 ( u + v ) t
A worked example A driver at 72 km/h 72\ \text{km/h} 72 km/h takes 0.8 s 0.8\ \text{s} 0.8 s to react, then the car slows at 6 m/s 2 6\ \text{m/s}^2 6 m/s 2 . How far does it go before it stops?
Units. 72 km/h = 20 m/s 72\ \text{km/h} = 20\ \text{m/s} 72 km/h = 20 m/s . Thinking distance. 20 m/s × 0.8 s = 16 m 20\ \text{m/s} \times 0.8\ \text{s} = 16\ \text{m} 20 m/s × 0.8 s = 16 m . Braking distance. ( 20 m/s ) 2 2 × 6 m/s 2 = 33.3 m \dfrac{(20\ \text{m/s})^2}{2 \times 6\ \text{m/s}^2} = 33.3\ \text{m} 2 × 6 m/s 2 ( 20 m/s ) 2 = 33.3 m . In all. 16 m + 33.3 m = 49.3 m 16\ \text{m} + 33.3\ \text{m} = 49.3\ \text{m} 16 m + 33.3 m = 49.3 m . Practice with new numbers
A program to try The program makes a table of stopping distances.
Read it, then run it.
Make the road wet: a = 4.0. How much longer is the stop at 90 km/h 90\ \text{km/h} 90 km/h ?
Mistakes to avoid Using the equations when the acceleration changes. Split the motion into parts. Forgetting the sign. Slowing down needs a negative a a a ; up and down need a chosen positive direction. The half in 1 2 a t 2 \tfrac12 at^2 2 1 a t 2 . Leave it out and the distance doubles. g g g at the top of a throw. The velocity is zero there, not the acceleration. 'Steady speed round a bend means no acceleration.' The direction changes, so it accelerates. Check yourself A car speeds up from 4 m/s 4\ \text{m/s} 4 m/s to 16 m/s 16\ \text{m/s} 16 m/s at 2 m/s 2 2\ \text{m/s}^2 2 m/s 2 . How long does it take? Stuck? Show a hint Show the answerA stone is dropped. How far does it fall in 2 s 2\ \text{s} 2 s ? ( g = 9.8 m/s 2 g = 9.8\ \text{m/s}^2 g = 9.8 m/s 2 ) Stuck? Show a hint Show the answerA ball is thrown up at 9.8 m/s 9.8\ \text{m/s} 9.8 m/s . How long until it reaches the top? Stuck? Show a hint Show the answerWhich equation finds a stopping distance when the time is not given? Stuck? Show a hint Show the answerA merry-go-round seat is 4 m 4\ \text{m} 4 m from the centre and goes round once every 8 s 8\ \text{s} 8 s . How fast does it move? Stuck? Show a hint Show the answer