Relative motion
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Interception
If one object is moving faster than another it may not be possible to set a collision course. However, in this section we deal only with problems where a collision is possible. If one object is to intercept another, then the relative velocity of one to the other lies along the straight line joining them.You may be asked to find the collision course. To solve the problem, begin by drawing a diagram showing the relative positions of the two objects. Then the line joining them is the line along which the relative velocity of one object to the other will lie. Draw a separate diagram showing the velocities of the two objects and their relative velocity acting along this line. The specific problem will then be soluble by substituting into the equation: AVB = VA - VB Example 1. A ship A lies 60 km to the west of a second ship B. A is travelling at 18 kmh-1 on a bearing of 52°. The ship B can travel at 22 kmh-1. Find the course that B should set in order to intercept A. Find also the time taken to interception. Solution Diagram of positions [Diagram goes here - download the original to see it.] The relative velocity of B to A lies along the line joining B to A; hence, the diagram of velocities is [Diagram goes here - download the original to see it.] Thus: B VA = VB - VA Let BVA = x i Then: -x i = -22 sin q i + 22 cos q j - (18 sin 52 i + 18 cos 52 j ) Resolving vertically: 0 = 22 cos q - 18 cos 52 [Equation goes here - download the original to see it.] q = 59.8° (0.1°) Bearing = 360 - 59.8 = 300.2° Resolving horizontally: [Equation goes here - download the original to see it.] [Equation goes here - download the original to see it.] Example 2. This illustrates the case where the resultant velocity does not lie along a "nice" horizontal or vertical A fighter plane, A, is traveling north at 1000 kmh-1. Another fighter, B, which can travel at 850 kmh-1 wishes to intercept it. It is currently positioned 1500km to the East and 2000km to the North of A. Find the course B should plot & the time taken to interception. Solution The following diagram shows the positions of the fighters. [Diagram goes here - download the original to see it.] The relative velocities will lie along the line joining B to A; hence, we calculate the bearing by [Diagram goes here - download the original to see it.] [Equation goes here - download the original to see it.] The diagram of the velocities is [Diagram goes here - download the original to see it.] The diagram shows that fighter B could intercept along two paths, with bearings q1 and q2. The shortest time will be given by the path with bearing q1. BVA = VA - VB Let |BVA| = x Then [Equation goes here - download the original to see it.] Note here we have put +850 cos q j because if q > 90 this will be a negative value, but we put -850 sin q i because we show that the horizontal component of VB is in the same direction as the horizontal component of BVA. Uncoupling: (Resolving horizontally and vertically separately): 1. [Equation goes here - download the original to see it.]2. [Equation goes here - download the original to see it.] From (1)Equation goes here - download the original to see it.] Substituting in (2) [Equations go here - download the original to see it.] The shortest approach is given by q1 = 98.23° Horizontal component = X sin 36.8 [Equation goes here - download the original to see it.]
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Contents of
Relative motion
1 Relative Motion
2 Interception
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