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

2. 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

3. Types of Questions Velocity Time Graph Gravity/Vertical Motion
Period of Constant Speed
No Period of Constant Speed
Gravity/Vertical Motion
2 Bodies in Motion/Overtaking
Passing successive Points/2 points

4. Section 1 Velocity Time Graphs

5. 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 t1 t2

6. 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

7. Area of whole shape is 30
Area = 6
Velocity v
Time 2t
6-3t t
Relate Times together
Area of acceleration = 6

8. Finding area of triangles and rectangle
Total Area = 30 m
Finding area of triangles and rectangle
Finding Distance (area of rectangle)

9. Example 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.

10. Ratios are easier as fractions
Velocity
0.6
0.8
Ratios are easier as fractions
Time t
14-t
Fact: Ratio of accelerations

11. Area of left triangle and area of right triangle
When Accelerating
Area of left triangle and area of right triangle
Total Distance = Total Area

12. 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.
Draw a speed time graph for the motion
Find d in terms of f and t.

13. Use info on acceleration Time t1 t2
Ratios of Times
Velocity f 3f
Total Area
Use info on acceleration
Time t1 t2
Told t in the question is total time
E. Williamson, IMTA Revision Seminar

14. Area of left Triangle
Area of right triangle

15. Now try some questions by yourself on the attached sheet

16. Section 2 Passing a number of points

17. 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

18. 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/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=1/3
The train comes to rest s metres after passing r. Find s, giving your answer correct to the nearest metre.

19. Start from same point each time i.e. pq r w
u = u
a = -f
125 m
125 m s
p to q
p to r
Start from same point each time i.e. p

20. Solving
p to w
Get total distance from p to w and then subtract distance p to r from it

21. 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

22. Fifth Second is between t=4 and t=5 Initial Velocity is u and acc=a
t=5, u=u, s=s, t=5, a=a
t=4, u=u, s=s, t=4, a=a
From Question
t=7, u=u, s=s, t=7, a=a
t=6, u=u, s=s, t=6, a=a
E. Williamson, IMTA Revision Seminar

23. Now try some questions by yourself on the attached sheet