Kinematics PH1.2

Motion – defining important terms: (1) distance and displacement scalar vector unit: m displacement distance symbol: s

… the same speed, but different velocities (2) speed and velocity scalar vector … the same speed, but different velocities

distance time average speed = unit: m/s or ms-1 Δs t v = ds dt from GCSE work … or average velocity in a straight line unit: m/s or ms-1 using symbols, we write … v = Δs t Δ means ‘a change in …’ If t is so small that there is not enough time for the speed to change … d means ‘a very small change in …’ ds dt instantaneous speed =

… is the rate of change of the velocity (3) acceleration not speed!! … is the rate of change of the velocity For motion in a straight line … change in velocity time taken average acceleration = a = Δv t unit: m/s2 or ms-2 in symbols … dv dt instantaneous acceleration = in a very small time …

Motion graphs higher constant speed (1) Displacement-time graph acceleration displacement constant speed at rest time constant speed in reverse

speed = gradient of graph s Calculating speed from a displacement-time graph (1) constant speed displacement speed = gradient of graph s time t speed = distance time s t =

ds dt (2) instantaneous speed displacement speed is constant during a small time interval ds enlarging a small part of the graph … dt time

ds dt ds dt instantaneous speed = = gradient of this section displacement ds dt time

s instant. speed t = s t ds dt instantaneous speed = = gradient of this line displacement s t instant. speed = s the gradient is the same, irrespective of the size of the triangle t time

change in instantaneous speed Calculating acceleration from a displacement-time graph change in instantaneous speed time taken to change acceleration = displacement (s2 / t2) - (s1 / t1) T2 – T1 acceleration = T1 T2 t2 s2 time taken to change t1 s1 time

Motion graphs larger constant acceleration (2) Velocity-time graph non-uniform acceleration constant acceleration velocity constant speed time constant deceleration, then constant acceleration in reverse

acceleration = gradient of the graph Calculating acceleration from a velocity-time graph velocity time v t acceleration = gradient of the graph acceleration = change in velocity time taken to change v t =

v t v = t dv dt instantaneous acceleration = = gradient of this line velocity instantaneous acceleration v t = v t time

Calculating the distance travelled … velocity v time t distance = speed × time = v·t distance = area under the graph

Calculating the distance travelled during a constant acceleration … velocity v time t distance = area under the graph = ½v·t

Calculating the distance travelled during a changing acceleration … velocity time distance = area under the graph

distance = area under the graph Calculating the distance travelled during a changing acceleration … velocity time distance = area under the graph

distance = area under the graph distance = A1 + A2 + A3 + A4 Calculating the distance travelled during a changing acceleration … A2 A1 A3 A4 distance = area under the graph distance = A1 + A2 + A3 + A4

The equations of motion only considering motion in a straight line constant acceleration only we use symbols … u – initial velocity v – final velocity s – displacement t – time a – acceleration

Equations of motion v = u + at By definition … change in velocity time taken acceleration = or in symbols … v – u t a = after rearranging … v = u + at

average velocity = ½(u+v) From the velocity-time graph for the journey … velocity v average velocity = ½(u+v) u time t distance travelled = average velocity × time s = ½(u+v)t

From the first eqn. of motion, v = u + at Substituting for v in the second eqn. of motion, s = ½(u+v)t s = ½(u +u + at)t s = ½(2u + at)t s = ½(2u + at)t s = ½(2ut + at2) s = ut +½at2 hence

v = u + at t = (v-u) / a 2as = uv + v2 – u2 – uv v2 = u2 + 2as From the first eqn. of motion, v = u + at so t = (v-u) / a Substituting for t in the second eqn. of motion … s = ½(u+v)t s = ½(u+v)(v-u) / a s = (u+v)(v-u) / 2a 2as = (u+v)(v-u) 2as = uv + v2 – u2 – uv 2as = uv + v2 – u2 – uv v2 = u2 + 2as hence

Projectiles a projectile is any object that moves solely under the influence of gravity we ignore the effects of air resistance we analyse motion in two dimensions

Dropping a projectile (i.e. freefall) … Galileo’s experiment in Pisa, the hammer and feather experiment on the Moon without air resistance, acceleration = 9.8ms-2 and is independent of the object’s mass

Launching a projectile horizontally … a steel ball is pushed sideways a steel ball is dropped multi-flash photos show that … vertical motion and horizontal motion are independent of each other

constant acceleration g = 9.8ms-2 downwards Vertical motion the distance between the lines is getting smaller gravity pulling it downwards constant acceleration g = 9.8ms-2 downwards Horizontal motion equal distance between lines no force acting constant speed sideways

Resolving a vector – splitting a vector into two components concentrate on two perpendicular vectors θ v y x vy = v·sinθ vx = v·cosθ

Launching a projectile at an angle uy = u·sinθ u first of all, the initial velocity (u) needs to be resolved into components θ vertical component of the velocity is changing throughout ux = u·cosθ horizontal component of the velocity is constant the projectile’s speed and direction is constantly changing

time to climb and fall = time available to travel sideways

force = mass × acceleration driving force G drag D force = mass × acceleration the acceleration is directly proportional to the magnitude of the resultant force G - D = m × a G - D acceleration a = m

When the driving force > drag, the car accelerates As the car accelerates, the drag increases and the resultant force decreases therefore the acceleration decreases, but the car is still accelerating, so … … this is repeated until driving force = drag no resultant force → no acceleration → terminal velocity

during freefall, resultant force = weight – air resistance On jumping, resultant force = mg and acceleration = g When he starts to fall, air resistance appears so the resultant force decreases and the acceleration < g … but he is still accelerating and air resistance continues to grow… so the resultant force and acceleration are still decreasing ...

during freefall, resultant force = weight – air resistance This is repeated, until the speed is large enough for air resistance = parachutist’s weight …. At this point, resultant force = 0 and acceleration = 0 …. The parachustist reaches a constant speed, i.e. the terminal velocity

during freefall, resultant force = weight – air resistance When the weight is small ….. only a small air resistance is needed to cancel the weight, so a high speed is not needed for the air resistance to grow sufficiently… therefore the terminal velocity is small