2Ordinary LevelBefore you start this course, it is highly recommended you complete the basics in the ordinary level course. This will give you the basics in the different sections
3Types of Questions Velocity Time Graph Gravity/Vertical Motion Period of Constant SpeedNo Period of Constant SpeedGravity/Vertical Motion2 Bodies in Motion/OvertakingPassing successive Points/2 points
5Velocity Time Graph Area under Curve = distance travelled Slope (y over x) is accelerationUse equations of motion for constant acceleration period only.avelocitytimedt1t2
6Example 1.1 [LC:1997 Q1(a)]A particle, moving in a straight line, accelerates uniformly from rest to a speed v m/s. It continues at this constant speed for a time and then decelerates uniformly to rest, the magnitude of the deceleration being twice that of the acceleration. The distance travelled while acceleration is 6 m. The total distance travelled is 30 m and the total time taken is 6 s.Draw a speed time graph and hence or otherwise, find the value of v.Calculate the distance travelled at v m/s
7Area of whole shape is 30Area = 6VelocityvTime2t6-3ttRelate Times togetherArea of acceleration = 6
8Finding area of triangles and rectangle Total Area = 30 mFinding area of triangles and rectangleFinding Distance (area of rectangle)
9Example 1.2 [LC: 1994 Q1(a)]A lift in a continuous descent, had uniform acceleration of 0.6 m/s2 for the first part of its descent and a retardation of 0.8 m/s2 for the remainder. The time, from rest to rest was 14 seconds. Draw a velocity time graph and hence, or otherwise, find the distance descended.
10Ratios are easier as fractions Velocity0.60.8Ratios are easier as fractionsTimet14-tFact: Ratio of accelerations
11Area of left triangle and area of right triangle When AcceleratingArea of left triangle and area of right triangleTotal Distance = Total Area
12Example 1.3 [LC: 2006 Q1(a)]A lift starts from rest. For the first part of its descent, it travels with uniform acceleration f. It then travels with uniform retardation 3f and comes to rest. The total distance travelled is d and the total time taken is t.Draw a speed time graph for the motionFind d in terms of f and t.
13Use info on acceleration Time t1 t2 Ratios of TimesVelocityf3fTotal AreaUse info on accelerationTimet1t2Told t in the question is total timeE. Williamson, IMTA Revision Seminar 2010
17Passing number of points Always start from same point to use same initial velocity i.e. if question is a to b, b to c and c to d; you treat it as a to b a to c a to d
18Example 2.1 [LC:2003]The points p, q, and r all lie in a straight line.A train passes point p with speed u m/s. The train is travelling with uniform retardation f m/s2.The train takes 10 seconds to travel from p to q and 15 seconds to travel from q to r, where |pq|=|qr|=125 metres.Show that f=1/3The train comes to rest s metres after passing r. Find s, giving your answer correct to the nearest metre.
19Start from same point each time i.e. p qrwu = ua = -f125 m125 msp to qp to rStart from same point each time i.e. p
20Solvingp to wGet total distance from p to w and then subtract distance p to r from it
21Example 2.2 [LC:1988 Q1 (a)]A particle moving in a straight line with uniform acceleration describes 23 m in the fifth second of its motion and 31 m in the seventh second. Calculate its initial velocity
22Fifth Second is between t=4 and t=5 Initial Velocity is u and acc=a t=5, u=u, s=s, t=5, a=at=4, u=u, s=s, t=4, a=aFrom Questiont=7, u=u, s=s, t=7, a=at=6, u=u, s=s, t=6, a=aE. Williamson, IMTA Revision Seminar 2010
23Now try some questions by yourself on the attached sheet