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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

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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

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**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

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**Section 1 Velocity Time Graphs**

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**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

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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

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

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**Finding area of triangles and rectangle**

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

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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.

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**Ratios are easier as fractions**

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

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**Area of left triangle and area of right triangle**

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

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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.

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**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 2010

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Area of left Triangle Area of right triangle

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**Now try some questions by yourself on the attached sheet**

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**Section 2 Passing a number of points**

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**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

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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.

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**Start from same point each time i.e. p**

q 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

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

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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

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**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 2010

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**Now try some questions by yourself on the attached sheet**

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