A boat travels in a straight path that is 25° west of north. Which describes the values of the west and north components of the
2 answers:
You might be interested in
Answer:
59.26°
Explanation:
Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.
Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.
Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration = acosθ.
So, y = ut + 1/2a't²
y = 0 × t + 1/2(acosθ)t²
y = 0 + 1/2(acosθ)t²
y = 1/2(acosθ)t² (1)
Also, both particles must move the same horizontal distance to collide in time, t.
Let x be the horizontal distance,
x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision
Also, using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration = asinθ.
So, x = ut + 1/2a"t²
x = 0 × t + 1/2(ainsθ)t²
x = 0 + 1/2(asinθ)t²
x = 1/2(asinθ)t² (3)
Equating (2) and (3), we have
vt = 1/2(asinθ)t² (4)
From (1) t = √[2y/(acosθ)]
Substituting t into (4), we have
v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²
v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)
v√[2y/(acosθ)] = ytanθ
√[2y/(acosθ)] = ytanθ/v
squaring both sides, we have
(√[2y/(acosθ)])² = (ytanθ/v)²
2y/acosθ = (ytanθ/v)²
2y/acosθ = y²tan²θ/v²
2/acosθ = ytan²θ/v²
1/cosθ = aytan²θ/2v²
Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1
secθ = ay(sec²θ - 1)/2v²
2v²secθ = aysec²θ - ay
aysec²θ - 2v²secθ - ay = 0
Let secθ = p
ayp² - 2v²p - ay = 0
Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have
ayp² - 2v²p - ay = 0
0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0
10.85p² - 15.68p - 10.85 = 0
dividing through by 10.85, we have
p² - 1.445p - 1 = 0
Using the quadratic formula to find p,
Since p = secθ
secθ = 1.95625 or secθ = -0.51125
cosθ = 1/1.95625 or cosθ = 1/-0.51125
cosθ = 0.5112 or cosθ = -1.9956
Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.
So, cosθ = 0.5112
θ = cos⁻¹(0.5112)
θ = 59.26°
So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.
Answer:
In the analytical method,
- Resolve the vectors into the perpendicular components of the Cartesian coordinates.
- Calculate the magnitude of the resultant vector using the Pythagoras theorem.
Explanation:
- There are two methods to find the magnitude of the resultant vector.
- One is the geometrical method and the other one is the analytical method.
- In the geometrical method, all the vectors are connected the head to tail with the appropriate magnitude and the resultant vector is obtained by joining the initial point and the final point by a vector in the reverse direction. The magnitude of the resultant vector is given by the length of the line.
- In the analytical method, all the vectors are resolved into the perpendicular components.
- Using Pythagoras theorem, the magnitude of the resultant vector can be obtained
- If A and B are the two vectors forming an angle ∅ between them, then the magnitude of the resultant vector is given by the formula