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TOPIC 8 Describing Straight Line Motion in 2D Space

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Elementary Vector Algebra in Two Dimensions SCALARS are quantities such as mass, volume, distance, speed and temperature which have a size but no associated direction. VECTORS are quantities such as displacement, velocity, force and acceleration which have both a size and an associated direction. A vector can be represented by a straight line whose length reflects the magnitude of the vector and whose orientation indicates the direction of the vector. (a) Directed Line Segment - this is a vector joining two points P Q QP O

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Elementary Vector Algebra in Two Dimensions (b) Position Vector - this is a vector joining a point to the origin The magnitude of a vector, a, is known as its MODULUS and written as a A UNIT VECTOR in a given direction is a vector with unit magnitude in that direction. The symbol i is used for the unit vector along the x axes and the symbol j is used for the unit vector along the y axes. When we write a vector using i and j it is said to be written in COMPONENT FORM. P O p

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Elementary Vector Algebra in Two Dimensions For example, consider the point L (4,7) It has the associated position vector l = 4i + 7j To find its modulus we use Pythagoras’ Theorem l = √x 2 + y 2 Definition of Modulus = √4 2 + 7 2 = √16 + 49 l = √65 L l = 4i + 7j 4 7

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Elementary Vector Algebra in Two Dimensions Example 1 (a) Change the column vector into unit vector form. (b) Convert the vector -7j into column vector form Answer 1 (a) = 5i + 8j (b) -7j =

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Elementary Vector Algebra in Two Dimensions Example 2 Given a = 3i + 2j and b = 5i – 6j find the resultant of a and b. Answer 2 a + b = 3i + 2j + 5i – 6j = 8i - 4j Example 3 If p = 2i – 8j and q = -9i + j work out 2p – 3q. Answer 3 2p – 3q = 2(2i – 8j) – 3(-9i + j) = 4i – 16j + 27i – 3j = 31i – 19j

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Elementary Vector Algebra in Two Dimensions Example 4 Forces F 1 = (6i + 4j)N, F 2 = (-2i – 5j)N and F 3 = (ai + bj)N are in equilibrium. Find the value of a and b. Answer 4 If F 1, F 2 and F 3 are in equilibrium then: F 1 + F 2 + F 3 = 0 so(6i + 4j) + (-2i – 5j) + (ai + bj) = 0 (6 – 2 + a)i + (4 – 5 + b)j = 0 (4 + a)i + (-1 + b)j = 0 So the i components and the j components must be zero 4 + a = 0-1 + b = 0 =>a = -4=>b = 1

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Using Vectors To Describe Motion in the Plane In Topic 1 we studied the constant acceleration formulae for motion in one dimension: v = u + at s = ut + ½at 2 s = ½(u + v)t Using vectors we can now use similar formulae to study motion in two dimensions: v = u + at s = ut + ½at 2 s = ½(u + v)t

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Using Vectors To Describe Motion in the Plane In Topic 3 we studied Newton’s Second Law for motion in one dimension: F = ma Using vectors we can now use a similar equation to study motion in two dimensions: F = ma Example 5 Initially a particle P of mass 3kg is at rest at the point D whose displacement relative to the origin O is OD = (i + 2j)metres. If the force F = (9i + 3j)newtons acts on P, find the displacement OE of P from O after 2 seconds.

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Using Vectors To Describe Motion in the Plane Example 6 A particle of mass 5kg is acted upon by a constant force of: (15i + 20j)newtons If, after 8 seconds, its velocity is: (24i + 20j)m/s Find its initial velocity.

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CHAPTER 3: VECTORS NHAA/IMK/UNIMAP.

CHAPTER 3: VECTORS NHAA/IMK/UNIMAP.

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