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Fofino [41]
3 years ago
10

Xinran walks 3 mi/h uphill, 4 mi/h on flat land and 5 mi/h downhill. If he walks one mile uphill, then one mile on flat land and

then returns by the same route to his starting point, how many minutes does he walk?
Mathematics
1 answer:
Dmitriy789 [7]3 years ago
7 0

Given :

Xinran walks 3 mi/h uphill, 4 mi/h on flat land and 5 mi/h downhill.

To Find :

If he walks one mile uphill, then one mile on flat land and then returns by the same route to his starting point, how many minutes does he walk.

Solution :

Time taken to walk one mile uphill is :

t_1 = \dfrac{1}{3}\ hours

Time taken to walk two mile on flat land( one on going and one or returning ) is :

t_2 = \dfrac{2}{4}\ hours\\\\ t_2 = \dfrac{1}{2}\ hours

Time taken to walk one mile downhill is :

t_3 = \dfrac{1}{5}\ hours

Total time taken :

T = t_1 + t_2 + t_3 \\\\T = \dfrac{1}{3} + \dfrac{1}{2}+\dfrac{1}{5}\\\\T = \dfrac{10 + 15 + 6}{30} \\\\T = \dfrac{31}{30}\ hours  = \dfrac{31}{30}\times 60  \ minutes\\\\T = 62\ minutes

Therefore, time taken un 62 minutes.

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g 1.32 Two points on a sphere of radius 3 are given as P1(3,0,30) and P2(3,45,45): (a) Find the position vectors of P1 and P2. (
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Answer:

a) P.V  of is OP₁ = [ 1.5i + 0j + 2.6k ],   P.V  of is OP₂ = [ 1.5i + 1.5j + 2.12k ]

b) Vector connecting P₁ to P₂ is [ 0i + 1.5j + 0.48k ]  

c) cylindrical coordinates are (1.5, π/2, 0.48)

Step-by-step explanation:

Given that;

r = 3

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a)

P.V of P₁

x = rcos∅sin∅ = 3(cos0°) ( sin30°) = (3 × 1 × 0.5) = 1.5

y = rsin∅sin∅  = 3(sin0°) (sin30°)   = (3 × 0 × 0.5) = 0

z = rcos∅        = 3(cos30°)             = ( 3 × 0.866)  = 2.6

∴ P.V  of is OP₁ = [ 1.5i + 0j + 2.6k ]

P.V of P₂

x = rcos∅sin∅ = 3(cos45°) ( sin45°) = (3 × 0.7071 × 0.7071) = 1.5

y = rsin∅sin∅  = 3(sin45°) (sin45°)   = (3 × 0.7071 × 0.7071) = 1.5

z = rcos∅        = 3(cos45°)                 = ( 3 × 0.7071)            = 2.12

∴ P.V  of is OP₂ = [ 1.5i + 1.5j + 2.12k ]

b)

Vector connecting P₁ to P₂ is given by

OP₂ - OP₁ = [ 1.5i + 1.5j + 2.12k ] - [ 1.5i + 0j + 2.6k ]

= [ 0i + 1.5j + 0.48k ]  

c)

P₁P₂ → = [ 0i + 1.5j + 0.48k ]  = [ 1.5j + 0.48k ]  

so in a cylindrical coordinate, it should be

r = √(o² + 1.5²) = 1.5

∅ = tan⁻¹[y/π] = π/2

z = 0.48

cylindrical coordinates are (1.5, π/2, 0.48)

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