Linear Accelerated Motion Part 1 For the Higher Level Leaving Cert Course ©Edward Williamson, Applied Maths Local Facilitator, Coachford College, Co Cork

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Linear Accelerated Motion Part 1 For the Higher Level Leaving Cert Course Edward Williamson, Applied Maths Local Facilitator, Coachford College, Co Cork Slide 2 www.pdst.ie Ordinary Level Before 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 Slide 3 www.pdst.ie Types of Questions Velocity Time Graph Period of Constant Speed No Period of Constant Speed Gravity/Vertical Motion 2 Bodies in Motion/Overtaking Passing successive Points/2 points Slide 4 www.pdst.ie Section 1 Velocity Time Graphs Slide 5 www.pdst.ie Velocity Time Graph Area under Curve = distance travelled Slope (y over x) is acceleration Use equations of motion for constant acceleration period only. a velocity time d t1t1 t2t2 Slide 6 www.pdst.ie Example 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 Slide 7 www.pdst.ie Velocity Time 2t t v 6-3t Area = 6 Area of whole shape is 30 Area of acceleration = 6 Relate Times together Slide 8 www.pdst.ie Total Area = 30 m Finding area of triangles and rectangle Finding Distance (area of rectangle) Slide 9 www.pdst.ie Example 1.2 [LC: 1994 Q1(a)] A lift in a continuous descent, had uniform acceleration of 0.6 m/s 2 for the first part of its descent and a retardation of 0.8 m/s 2 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. Slide 10 www.pdst.ie Velocity Time t14-t 0.6 0.8 Fact: Ratio of accelerations Ratios are easier as fractions Slide 11 www.pdst.ie When Accelerating Total Distance = Total Area Area of left triangle and area of right triangle Slide 12 www.pdst.ie Example 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. i. Draw a speed time graph for the motion ii. Find d in terms of f and t. Slide 13 www.pdst.ie f 3f3f Velocity Time t1t1 t2t2 Ratios of Times Total Area Told t in the question is total time Use info on acceleration Slide 14 www.pdst.ie Area of left Triangle Area of right triangle Slide 15 www.pdst.ie Now try some questions by yourself on the attached sheet Slide 16 www.pdst.ie Section 2 Passing a number of points Slide 17 www.pdst.ie Passing 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 Slide 18 www.pdst.ie Example 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/s 2.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. i) Show that f=1/3 ii) The train comes to rest s metres after passing r. Find s, giving your answer correct to the nearest metre. Slide 19 www.pdst.ie 125 m pq r u = u a = -f p to q p to r Start from same point each time i.e. p w s Slide 20 www.pdst.ie Solving p to w Get total distance from p to w and then subtract distance p to r from it Slide 21 www.pdst.ie Example 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 Slide 22 www.pdst.ie Fifth Second is between t=4 and t=5 Initial Velocity is u and acc=a t=4, u=u, s=s, t=4, a=a t=5, u=u, s=s, t=5, a=a From Question t=7, u=u, s=s, t=7, a=a t=6, u=u, s=s, t=6, a=a Slide 23 www.pdst.ie Now try some questions by yourself on the attached sheet