KINEMATICS

==Kinematic Equations== WHen the acceleration of an object is constant, the following set of equations can be used to describe and predict the object's motion. $$s = \frac {(u + v)}{2} \times t$$ $$v = u + at$$ $$ s = ut + ½at^2$$ $$ v^2 = u^2 + 2as $$ t = time a = acceleration v = final velocity u = initial velocity s = displacement Sometimes displacement has the symbols '''d''', '''x''', or '''y''' depending on the convention used. Similarly, initial velocity is sometimes written as v~I~ and final velocity as v~F~. ==Example Question== A car begins to move from rest with a constant acceleration of a = 5.0 ms^-2^. How far has it moved after 4.0s in the direction of acceleration? [image:http://i.imgur.com/dCaMrkz.png] '''Define the space:''' We will define the direction of the acceleration at the + direction. List the quantities we know and want to find: u = 0 (initial velocity) a = +5.0ms^-2^ t = 4.0s s = ? Choose an equation which includes all the quantities listed: $$ s = ut + ½at^2$$ Plug the values into the equation including the correct direction sign for each quantity: $$s = (0 \times 4.0) + ½ \times 5.0 \times 4.0^2$$ $$s = 40m$$ ==Example Question== An object slides without friction up a steady slope with an initial velocity of 4.0ms^-1^ at t = 0. It takes 16m for the object to come to rest. Calculate the object’s acceleration and the time it took to come to rest. [image:http://i.imgur.com/cMYqFN5.png] '''Define the space:''' + is up the slope - is down the slope. Therefore we expect the acceleration to have a '''negative''' value when calculated at the object is slowing down s = +16m (up the slope) v = 0 u = +4.0ms^-1^ (up the slope) a = ? t = ? v^2^ = u^2^ + 2as a=(v^2^- u^2^)/2s a=(0^2^ - 4.0^2^)/(2 x 16) '''a = -0.50ms^-2^''' then v =u+at t = (v-u)/a t = (0-4.0)/(-0.50) '''t =8.0s''' ==Example Question== An object is in free-fall, falling from rest near the surface of the Earth. What equation would be most useful when making simple measurements of height of the drop? $$ s = ut + ½at^2$$ simplifies to $$ s = ½at^2$$ So taking a as 9.81ms^-2^ or 10ms^-2^ and knowing the time, you can quickly calculate the height of the drop.