Question

a golfer takes three putts to get the ball into the hole.the first put displaces the ball 3.66m north,the second 1.83m south east and the third 0.99m south West. what is it's magnitude?

Answer 1

this question can be easily solved by $\underline{\text{vector method}}$ .
See attachment, A rough daigram is shown in figure. Golfer first displaces 3.66 m north ,so in vector d₁ = 3.66 j . second displaces 1.83 south - east. In vector form d₂ = 1.83cos45° i - 1.83sin45° j and finally displaces 0.99m south- west , in vector form , d₃ = -0.99cos45° i - 0.99sin45° j

We have to find magnitude of displacement ?
displacement = d₁ + d₂ + d₃
= 3.66j + 1.83cos45°i - 1.83sin45°j - 0.99cos45°i - 0.99sin45°j
= 3.66j + (1.83 - 0.99)cos45°i - (1.83 + 0.99)sin45°j
= 3.66j + 0.84× 1/√2 i - 2.82 × 1/√2j
= 3.66j + 0.6i - 2j
= 1.66j +0.6i

Now, magnitude of displacement = |1.66j +0.6i| = √(1.66² + 0.6²) ≈ 1.76 m

Answer 2

Given Conditions ⇒
Golfer displaces the ball 3.66 m in north direction.
∴ Vector in this case (a₁) = 3.66 j.

Now, Golfer displaces the ball 1.83 m in South East direction.
∴ Vector in this case (a₂) = 1.83 Cos 45° - 1.83 Sin 45° j.
 
Golfer again displaces the ball 0.99 m in the South west direction.
∴ In Vector form (a₃) = -0.99 Cos 45° - 0.99 Sin 45° j.

Refers to the attachment for the better understanding.

Now,
∵ Total Displacement (a) = a₁ + a₂ + a₃
 = 3.66 j + 1.83 Cos 45°i - 1.83 Sin 45° j - 0.99 Cos 45° i - 0.99 Sin 45° j.
 = 3.66 j + 1.83 Cos 45° i - 0.99 Cos 45° i - 1.83 Sin 45° j - 0.99 Sin 45° j.
 = 3.66 j + Cos 45°i (1.83 - 0.99) - Sin 45°j (1.83 + 0.99)
 = 3.66 j + Cos 45°i (0.84) - Sin 45° j(2.82)
[ ∵ Cos 45° = Sin 45° = 1/√2]

 = 3.66 j + 1/√2 × 0.84 i - 1/√2 × 2.82 j
 =  1.66 j + 0.6 i

∴ Magnitude of the Displacement = $\sqrt{1.66j + 0.6 i}$
 = $\sqrt{(1.66)^{2} + (0.6)^{2} }$
 = $\sqrt{2.7556 + 0.36}$
 = $\sqrt{3.1156}$
 = 1.765 m.


Hence, the magnitude of the Displacement covered by the ball is 1.756 m.


Hope it helps.