(a) A(-1, 2), B(3, 5) and C(4, 8) are the vertices of triangle ABC. Forces whose magnitudes are 5N and (3sqrt{10})N act along (overrightarrow{AB}) and (overrightarrow{CB}) respectively. Find the direction of the resultant of the forces.
(b) A particle starts from rest and moves in a straight line. It attains a velocity of 20 m/s after covering a distance of 8 metres. Calculate :
(i) its acceleration ; (ii) the time it will take to cover a distance of 40 metres.
Explanation
(a)
(|AB| = sqrt{(-1 – 3)^{2} + (2 – 5)^{2}})
= (sqrt{16 + 9} = sqrt{25})
= 5 units.
(|AC| = sqrt{(4 – 1)^{2} + (8 – 2)^{2}})
= (sqrt{25 + 36})
= (sqrt{61}) units.
(|BC| = sqrt{(4 – 3)^{2} + (8 – 5)^{2}})
= (sqrt{1 + 9} )
= (sqrt{10}) units.
From the cosine rule,
((sqrt{61})^{2} = 5^{2} + (sqrt{10})^{2} – 2(5)(sqrt{10}) cos <PBQ)
(cos < PBQ = frac{25 + 10 – 61}{2(5)(sqrt{10})})
(cos < PBQ = frac{-26}{10sqrt{10}} = -0.4747)
(< PBQ = 118°, beta = < PBQ = 118°)
(< BPS = alpha = 180° – 118° = 62°)
The magnitude of the resultant
(|R|^{2} = 5^{2} + (3sqrt{10})^{2} – 2(5)(3sqrt{10}) cos 62)
= (25 + 90 – (30sqrt{10})(0.4695))
= (115 – 44.538)
= (70.462)
(|R| = sqrt{70.462} = 8.39N)
Let (delta) be the angle the resultant makes with the 5N force.
(PS^{2} = BS^{2} + BP^{2} – 2(BS)(BP) cos PBS)
((3sqrt{10})^{2} = (sqrt{70.462})^{2} + 5^{2} – 2(8.39)(5) cos delta)
(90 = 70.462 + 25 – 83.9 cos delta)
(cos delta = frac{70.462 + 25 – 90}{83.9})
(cos delta = 0.0651)
(delta = cos^{-1} (0.0651) = 86.27°)
(approxeq 86°).
The resultant makes an angle of 86° with the 5N force.
(b)(i) (u = 0 ms^{-1} ; v = 20 ms^{-1} ; s = 8 m ; a = ?)
(v^{2} = u^{2} + 2as )
(20^{2} = 0^{2} + 2(8)a implies 400 = 16a)
( a = 25 ms^{-2})
(ii) (t = ?)
(s = ut + frac{1}{2} at^{2})
(40 = 0 + frac{1}{2} (25)(t^{2}))
(80 = 25t^{2})
(t^{2} = frac{80}{25} = 3.2)
(t = sqrt{3.2} = 1.79 s)
It takes 1.79s to cover a distance of 40m.