Deriving the equations Kinematics Deriving the equations

Objects moving under acceleration diagrams

Starter: covert these measures KINEMATICS 1 KUS objectives BAT derive the suvat equations BAT apply them to solve problems BAT change between velocity measures Starter: covert these measures 16 m/s to km/h 40 mph to km/h 3

Objects moving under acceleration Imagine a particle moving in a straight line with constant (uniform) acceleration What other measures should we consider in our model? a acceleration t time Key Q: what do we mean by constant acceleration?; what is the difference between distance and displacement (s) u initial velocity v final velocity s displacement What are typical units for these measures?

First uniform acceleration equation (1)

First uniform acceleration equation WB1 A train starts from rest at a station and moves with constant acceleration. 35 seconds later it passes a signal box with a speed of 50 ms-1 What is its acceleration in ms-2 ? 𝑠= u = v = a = t = 50=0+𝑎(35) 𝒗=𝒖+𝒂𝒕 𝑎= 50 35 = 10 7 𝑚𝑠 −2

Second uniform acceleration equation (2)

Second uniform acceleration equation example WB2 A boat accelerates from 4 ms-1 to 6.5 ms-1 over 8 seconds How far has the boat gone in this time? 𝑠= u = v = a = t = 𝒔= 𝒖+𝒗 𝟐 𝒕 𝑠= 4+6.5 2 80 𝑠= 42 m

Third uniform acceleration equation and (3) Key Q: Given first two eqns how can we use substitution to derive eqn 3 and 4?;

Third uniform acceleration equation example WB3 A space shuttle travelling at 20 kms-1 slows with a deceleration of 45 ms-2 for 2 minutes. How many km will it have travelled? 𝑠= u = v = a = t = 𝒔=𝒖𝒕+ 𝟏 𝟐 𝒂 𝒕 𝟐 𝑠=20000×120+(−45) 120 2 𝑠=1752000 𝑠=1752 𝑘𝑚

Fourth uniform acceleration equation and so (4)

Changing units of speed Change these measures to metres per second or km per hour 20 km/h → 30 ms-1 → Key Q: what steps to convert kmph to mph or ms-1 360 km/h →

Fourth uniform acceleration equation example WB4 A car is following a Lorry going at 60 kmph on the motorway. The car constantly accelerates at 3 ms-2 to overtake the lorry and keeps accelerating for the 120 metres it takes to overtake. How fast is the car travelling once it has overtaken the lorry? Give your answer in kmph. Does this seem likely for a real life model? 𝑠= u = v = a = t = 𝒗 𝟐 = 𝒖 𝟐 +𝟐𝒂𝒔 𝑣 2 = 60×1000 3600 2 +2 3 (120) 𝑣 2 = 8980 9 ≅998 𝑣= 998 ×3600 1000 ≅114 𝑘𝑚𝑝ℎ 60 𝑘𝑚𝑝ℎ is 37.5 𝑚𝑝ℎ and 114 𝑘𝑚𝑝ℎ is 71.25 𝑚𝑝ℎ

Summary If you know three of these values you can always use the equations to work out the other two

𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑠=20×10+(−4) 10 2 𝑠=−200 𝑚 𝑠= u = v = a = t = WB 5  An object decelerates uniformly at 4 ms-2 along a straight horizontal road. If the object travels for 10 seconds and starts at  20ms−1   Find the displacement of the object Describe the journey the object travels and find what distance it travelled 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑠= u = v = a = t = 𝑠=20×10+(−4) 10 2 𝑠=−200 𝑚 The object slowed to a stop then travelled back, passing where it started and ending up 200m behind where it started

WB 6  A motorcycle accelerates uniformly along a straight horizontal road so that, when it has travelled 30 metres, its velocity has increased from 15ms−1 to 18ms−1.  Find the acceleration of the motorcycle Find the time that it takes for the motorcycle to travel this distance. 𝑠= u = v = a = t =

𝑠=𝒖𝑡+ 1 2 𝑎 𝑡 2 𝑠= 𝑣−𝑎𝑡 𝑡+ 1 2 𝑎 𝑡 2 𝑠=𝑣𝑡 −𝑎 𝑡 2 + 1 2 𝑎 𝑡 2 WB 7 The FIFTH equation  There is a fifth equation of motion: 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 Can you derive this from the other equations? 𝑠=𝒖𝑡+ 1 2 𝑎 𝑡 2 𝑠= 𝑣−𝑎𝑡 𝑡+ 1 2 𝑎 𝑡 2 𝑠=𝑣𝑡 −𝑎 𝑡 2 + 1 2 𝑎 𝑡 2 𝑠=𝑣𝑡 − 1 2 𝑎 𝑡 2

𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 160=28𝑡− 1 2 (2) 𝑡 2 s= 160 u = v = 28 a = 2 WB7 The FIFTH equation  A cyclist reaches a straight stretch of road, of length 160 m, and travels its length with constant acceleration of 2 𝑚𝑠 −2 . If the cyclist reaches the end of the straight stretch at velocity 28 𝑚𝑠 −1 , how long could it have taken? 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 160=28𝑡− 1 2 (2) 𝑡 2 s= 160 u = v = 28 a = 2 t = 𝑡 2 −28𝑡+160=0 (𝑡−20)(𝑡−8)=0 𝑡=20 seconds or 𝑡= 8 seconds

𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 22=2𝑢+2𝑎 (1) 126=6𝑢+18𝑎 (2) WB8 A car is moving along a straight horizontal road with constant acceleration. There are three points A, B and C, in that order, on the road, where AB = 22m and BC = 104 m The car takes 2 secs to travel from A to B and 4 secs to travel from B to C Find The acceleration of the car The speed of the car at the instant it passes A From A to C s= 126 u = u v = a = a t = 6 From A to B s= 22 u = u v = a = a t = 2 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 22=2𝑢+2𝑎 (1) 126=6𝑢+18𝑎 (2) Question from 2017 AS assessment materials 126=6(11−𝑎)+18𝑎 Combine (1) and (2) 126=66+12𝑎 𝑎=5 𝑚𝑠 −2 Substitute to (1) 𝑢=6 𝑚𝑠 −1

remember for next lesson …. KINEMATICS 1 KUS objectives 1 derive the suvat equations 2 apply them to solve problems 3 change between velocity measures One thing I will do For the next lesson …. One thing I need to remember for next lesson …. One thing I understood well is …. Plenary pyramid

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