Peter's Physics Pages

Bridging Course - Lecture 4 (Vector Example problems)

Example problems. Fully worked answers are provided.

17. Find the length and direction of a displacement in the x-y plane from (1,2) to (6,14).

---------- The length of the displacement is given from Pythagoras' theorem, which is the square on the hypotenuse equals the sum of the squares on the other two sides. Length = √(52 + 122) = 13 The angle is given from the definition of the tangent:tanθ = 12/5 ∴ θ = +67.40 (maths convention)

18. Find the resultant displacement from adding:-
(a) a displacement 2 cm East and a displacement 3 cm North-West.
(b) a displacement 8 cm East and a displacement 12 cm North-West.
Compare the results of (a) and (b) and state a general result about adding a pair of vectors which are the same multiple of another pair.

---------- (a) Changing to approved units and using Cartesian coordinates, the resultant is: Ra = (20 - 30·cos450, 0 + 30·sin450)     = (-12, 21.2) mm The angle is given from the definition of the tangent:tanθ = 12/5 ∴ θ = +67.40 (maths convention) (b) Rb = (80 - 120·cos450, 0 + 120·sin450)     = (-48, 84.8) mm The second pair of displacements are both four times larger, so the resultant will be four times larger. This could be written: 4P + 4Q = 4(P + Q) = 4R

19. A ship sails 10 km ENE. Find the North and East components of its displacement.

---------- ENE is halfway between NE and E, i.e. 22.50 East of NE. The N component is 10·sin22.50 = 3.8 km The E component is 10·cos22.50 = 9.2 km

20. A rifle pointed SW and up at an angle of 35° from the horizontal fires a bullet which leaves the 0.7 m barrel at 500 m.s-1. Find the average acceleration in the rifle barrel.

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The x, y, z axes form a right handed system. This means that if you hold a screwdriver in your right hand and turn it to rotate a screwhead clockwise, then the screw moves away from you and into the material. An anti-clockwise (counter-clockwise) rotation moves the screw towards you. Check that you can see that looking down onto the compass plane rotating from the E (or x axis) to the N (or y axis) the rotation is anti-clockwise, so the z axis is upwards from the compass plane. Δv = v = a·t Δs = s = ½a·t2 Eliminate t by substituting t = v/a s = ½·a·(v/a)2 = ½·(v2/a) a = ½·(v2/s) a = ½·(5002/0.7) = 1.78×105 m.s-2. This will be in the direction along the barrel, i.e. a = [a·cos(-450), a·sin(-450), a·sin(+350)] a = [126, -126, 102] km.s-2

21. A person wishes to row across a river 500 m wide. The boat is pointed directly across the river and rowed at 2 km.hr-1 relative to the water. The current is 4 km.hr-1 relative to the river bank. Find
(a) how long it takes to cross the river.
(b) where the person lands in relation to the start.

---------- The resultant speed of the boat will be increased by the current down the river. However, as the current is at right angles to the speed across the river, there is no increase or decrease to the speed across the river. Speed across river = width of river/time to cross Re-writing: Time to cross = width of river/speed across river Time = 0.5/2 = 0.25 h or 15 minutes Similarly, as the boat speed across the river is at right angles to the current down the river, there is no increase or decrease to the speed down the river. 4 km/h downstream for 1/4 h gives a landing distance of 1 km downstream.

Note that the resultant velocity of the boat is 4.47 km/h 63.40 clockwise from the boat direction. The resulant displacement is given by 500/cos63.40 = 1117 m. When similar vectors are added at right angles their effects can be treated separately.

22. A watch has a second hand which is 20 mm long. Find
(a) the speed of the tip of the second hand.
(b) the velocity of the second hand at 0 s and 15 s.
(c) the change in velocity between 0 s and 15 s.
(d) the average (vector) acceleration between 0 s and 15 s.

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(a) Speed = (distance travelled)/(time taken) = circumference/60s = 2πr/60 = 2π·20/60 mm.s-1 = 2mm.s-1

 (b) v1 = Velocity at 0s = 2mm.s-1 in x direction. (b) v2 = Velocity at 15s = 2mm.s-1 in -y direction. (c) Change in velocity (0s -> 15s) is done by adding the negative direction. Δv = √(22 + 22) = 2.8 mm.s-1 in the direction of the centre of the circle. (d)    a = (Change in velocity)/(The time during which the speed changed) a = 2.8/15 = 0.19 mm.s-2 in the direction of the centre of the circle. In this case the acceleration produces a change in direction without a change in speed, as it always acts at right angles to the velocity.

 23. A ship is travelling due east at 18.6 km/h. At 10:00 am the navigator measures the angle to a lighthouse as 300 off the port bow. Thirty minutes later the angle is measured to be 600 off the port bow. Find the position of the ship at 10:30 am relative to the light house. ----------

The exterior angle of a triangle is equal to the sum of the interior opposite angles.
In the diagram there is an exterior angle of 600, and as one of the interior opposite angles is 300, then the other one must be 300. The triangle is therefore an isosceles triangle and the two sides must be equal in length.
This means that the distance the ship travels during that half-hour equals the distance from the lighthouse at the end of the half-hour. The distance travelled is 18.6×0.5 = 9.3km. At 10:30 am the ship is 9.3km from the lighthouse in a direction S 300 W.

24. An aeroplane's destination is 298 km East of its starting position. There is a steady wind from the North West at 60 km.hr-1. The pilot wishes to make the trip in 40 minutes. Find
(a) the direction the pilot should head towards.
(b) the airspeed that should be flown.

---------- The final velocity needs to be 298/(2/3) = 447 km/h due East. Adding the plane's velocity to the wind velocity is to give the final velocity. The plane's velocity needs to be at some angle N or E. Using the cosine rule: v2 = 602 + 4472 - 2×60×447×cos45 i.e. v = 407 km/h Using the sine rule: sinθ/60 = sin45/407 which gives sinθ = 0.1 θ = E 60 N

E1. A ship sets out to sail to a point 240 km due North. A storm suddenly blows the ship to a point that is 100 km due East of the starting point. Find the diplacement for the ship to now sail to its original destination.

---------- r = (-100, +240) km in Cartesian coordinates. r = √( 1002 + 2402 ) = 260km θ = tan-1(+240/-100) = -67.40 The negative angle indicates that the angle is in the second quadrant, i.e. it is (180 - 67.4) = 112.60 In Polar coordinates r = ( 260km, 112.60 )km email Write me a note if you found this useful