Chp 2 Kinematics Summary

2 Kinematics Term Definition SI unit & Symbol Formula Remarks/Examples 2.1 Speed, v Is the rate of change of distance m/s A car moving round a circular track takes 240 s to do a lap of 8 km. What is the speed of the car in (a) km/h (b) m/s Velocity, v Is the rate of change of displacement Displacement is distance travelled in a specified direction. m/s A car has a velocity of 72 km/h. How far does it travel in half a minute ? time distance speedAverage = time ntdisplaceme velocityAverage = 33.3m/ss 240 m 8000 time distance speedAverage (b) km/h 120 h3600 240 km8 time distance speedAverage (a) === === m 600602 1 s 3600 m 000 72 minute2 1 x km/h 72 time x velocity ntDisplaceme time ntdisplaceme velocity == = = = xx

Quantities Definition SI unit & symbol Formulae Worked Examples/Remarks Acceleration, a Is the rate of change of velocity m/s2 where a is the acceleration in m/s2 v is the final velocity in m/s u is the initial velocity in m/s t the time in seconds Rearranging the above equation, v = u + at Worked example 1 A car traveling at 10 m/s accelerates uniformly at 2 m/s2. Find its velocity in 5 s time. Solution Initial velocity, u = 10 m/s acceleration, a = 2 m/s2 time, t = 5 s v = u + at v = 10 + 2 x 5 = 20 m/s Worked example 2 A train slows from 20 m/s with a uniform deceleration of 2m/s2. How long will it take to reach 5 m/s ? Solution Initial velocity, u = 20 m/s Final velocity, v = 5 m/s Deceleration = 2 m/s2 Meaning acceleration = − 2m/s2 Rearranging the equation, changethe make to takentime change velocity nAccleratio = t uv a −= s 7.52 15 2 205 a uv t =− −=− −= −= t uv a −=

2.2 Graphical Analysis of motion Displacement-time graph / distance-time graph Velocity-time graph Speed-time graph Acceleration-time graph Remarks 1. The gradient of displacement-time graph tells us about the velocity. 2. The gradient of the velocity-time graph tells us the acceleration. 3. The area under the velocity-time graph tells us the distance travelled. Both (i) and (ii) show that the object is at rest, not moving Gradient is zero, distance not changing with time. (i) shows object at rest, Speed , v = 0. (ii)shows object moving at constant or uniform speed. gradient=acceleration = zero. (i) Particle not accelerating. Acceleration is zero. (ii) Particle moves with constant acceleration A very important concept here is the gradient of a graph. It tells us how the vertical quantity changes with the horizontal quantity. Constant gradient shows constant speed. Constant gradient (or slope) shows constant acceleration. The acceleration is increasing at a constant rate. 0 t v i ii 0 t s displacement 0 t s i ii displacement velocity Δt Δs gradient = = 0 t v onaccelerati Δt Δv gradient = = velocity 0 t a i ii 0 t a

(i) decreasing gradient shows decreasing speed (ii) increasing gradient shows increasing speed (i) decreasing gradient shows decreasing acceleration (ii) increasing gradient shows increasing acceleration Object returns to starting point (i) at decreasing speed (ii) at increasing speed The speed is decreasing (i) at decreasing rate (ii) at increasing rate 0 t s i ii 0 t v i ii 0 t v i ii 0 t s i ii displacement velocity

This is the typical velocity- time graph showing (i) a train traveling between 2 stations (ii) A lift moving between two floors. As it leaves one floor for another, it accelerates until it reaches midway between the 2 floors whereby it starts to decelerate until it reaches the other floor. (i) This velocity-time graph shows a puck accelerating at an increasing rate. This is the typical velocity- time graph showing (i) an object falling from a great height such as from an aeroplane. (ii) It accelerates at a decreasing rate because of air resistance which increases as speed increases, causing the net force and hence the acceleration to decrease to zero. (iii)It could also represent an object falling through a viscous liquid. (i) This is the typical velocity-time graph showing an object thrown vertically upwards with a certain speed. (ii) It travels with uniform deceleration due to gravity. (iii) velocity = 0 when the object reaches its greatest height. O v t O t v Terminal velocity is the final velocity reached before a = 0 m/s2 t v O Object thrown vertically upwards Object at its greatest height when v = 0 m/s Object accelerates uniformly from rest and decelerates uniformly to rest travelling between 2 floors or 2 train stations v t Object falling from a great height through air or through a viscous liquid such as glycerine

2.3 Free-fall The 4 equations of linear motion ( motion in a straight line) Experiment shows that a body falling freely under gravity, accelerates uniformly at 10 m/s2 , This means that its velocity increases by 10 m/s every second of its fall. An object thrown vertically upwards decelerates at 10 m/s2 . This means its velocity decreases by 10 m/s every second as it rises until it reaches its maximum height. Thereafter it will accelerate uniformly as it falls freely under gravity. where v is the final velocity u is the initial velocity s is the distance travelled a is the acceleration The 4 equations above are the basic relations for motion under uniform acceleration. To show the first equation of motion To show the second equation of motion Distance travelled is given by area under the velocity-time graph To show the second equation of motion at u v u v at t uv a on,accelerati of definition By += −= −= 2asuv uv2as 2a uv a uv 2 uvs time x speedaverage travelledDistance 22 22 22 += −= −=      −     += = 2at2 1uts t2 at2us t2 u atu t2 uvs time x speedaverage travelledDistance +=      +=      ++=      += = 1) v = u + at 2) s = ½ (u + v) t 3) s = ut + ½ at2 4) v2 = u2 +2as

The two worked examples on the right show how the equations of motion can be used to solve kinematics problems. Worked example 1 on linear motion A ball is thrown vertically upwards from the ground with a velocity of 20 m/s. Calculate (a) the maximum height reached (b) the time to reach the maximum height (c) the time to reach the ground again after the ball is thrown up (d) the velocity reached half-way to the maximum height. (Assume acceleration due to gravity, g = 10 m/s2.) Solution (a) u = 20 m/s ; v = 0 m/s ; a = 10 m/s2 v2 = u2 + 2as 202 = 02 + 2 x 10 x s 400 = 20 s s = 400/20 = 20 m The maximum height reached is 20 m (b) By definition of acceleration, The time to reach the maximum height is 2 seconds. (c) The time to reach the ground again after the ball is thrown up is also 2 seconds. (d) v2=u2 + 2as = 202 − 2x10x10= 400−200 Worked example 2 on linear motion A car travels with a velocity of 18 km/h. It then accelerates uniformly and travels a distance of 50 m. If the velocity reached is 54 km/h, find the acceleration and the time to travel this distance. Solution Initial velocity, u = 18 km/h =18 000m/3600 s = 5 m/s Distance travelled = 50 m Final velocity , v = 54 km/h = 54 000m/3600s = 15 m/s v2 =u2 + 2as 152 = 52 + 2 x a x 50 200 = 100a a = 200/100 = 2 m/s2 The acceleration is 2 m/s2 s 210 200 a uvt t uva =− −= −= −= 14.1m/s200v == s 52 515 a uvt t uva =−= −= −=

(a)Describe in words what is happening