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Lesson Objective: Kinematics with Non Constant Acceleration WE CAN NO LONGER USE THE ‘SUVAT’ EQUATIONS !!

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Presentation on theme: "Lesson Objective: Kinematics with Non Constant Acceleration WE CAN NO LONGER USE THE ‘SUVAT’ EQUATIONS !!"— Presentation transcript:

1 Lesson Objective: Kinematics with Non Constant Acceleration WE CAN NO LONGER USE THE ‘SUVAT’ EQUATIONS !!

2 Instead we use calculus:

3 Example 1 : A particle moves in a straight line such that its displacement form a fixed point O is given by s = 4t 2 – t 3. Find a) s when t = 3. b) v when t = 2. c) a when t = 1.

4 Example 2 : The displacement of an particle from a fixed origin is given by: S = t 3 – 7.5t 2 + 18t – 2 a) Find the times and displacements of the particle when it is at instantaneous rest. b) Find the distance travelled by the particle in the first 5 seconds. c) Find the time when no resultant force is acting on the particle.

5 By reversing these results we can also say that: What will the definite integrals: represent?

6 Example 1 : A particle moves in a straight line with acceleration Initially, the particle is at the origin and moving with velocity 6ms -1. a) Find the velocity of the particle at time t = 4. b) Find the displacement of the object at time t = 4.

7 Lesson Objective General motion in 2d Consider a rock being thrown from the top of a cliff 10m tall at 20ms -` at an angle of 30 o to the horizontal. Assume that g = 10ms -1. a) What is the position of the rock relative to the foot of the cliff after 2 seconds? b) What is the position of the rock relative to the foot of the cliff after ‘t’ seconds? Vertically position from the foot of the cliff is y = 10 + 20sin30t- 5t 2 Horizontally position is x = 20cos30t We can wrap this up into one vector equation:

8 An example with non-constant acceleration An aircraft is dropping a crate of supplies on to level ground. Relative to an observer on the ground, the crate is released at the point with position vector and with initial velocity Its acceleration is modelled by a = for t ≤ 12 seconds i)Find an expression for the velocity of the crate at time ‘t’. ii)Find an expression for the position of the crate at time ‘t’. iii)Verify that the crate hits the ground 12s after its release and find how far from the observer the happens?

9 An example with different vector notation Relative to an origin on a long, straight beach, the position of a speedboat is modelled by the vector: r = (2t +2)i + (12 – t 2 )j where i and j are unit vectors perpendicular and parallel to the beach. i)Calculate the distance of the boat from the origin, O, when the boat is 6m from the beach ii)Sketch the path of the speedboat for t between 0 and 3 seconds (inclusive) iii)Find an expression for the velocity and acceleration of the speedboat at time t. Is the boat ever at rest? iv)For t = 3, calculate the speed of the boat and its angle and direction of motion to the line of the beach. v)Why is this model unrealistic if t is very large?

10 An example with forces A force of 12i +3tj N, where t is the time in seconds, acts on a particle of mass 6kg. The directions of i and j correspond to east and north respectively. i)Find a vector expression for the acceleration of the particle at time t. ii)Find the acceleration and its magnitude when t = 12. iii)At when time is the acceleration directed north-east? iv)If the particle starts with a velocity of 2i – 3j ms -1 when t = 0, what will its velocity be when t = 3?


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