Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
Answer:
Bank angle = 35.34o
Explanation:
Since the road is frictionless,
Tan (bank angle) = V^2/r*g
Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2
Tan ( bank angle) = 40^2/(230*9.81)
Tan (bank angle) = 0.7091
Bank angle = tan inverse (0.7091)
Bank angle = 35.34o
Jogger moves in three displacements
d1 = 10 blocks East
d2 = 5 blocks South
d3 = 2 blocks East
now we can say
total displacement towards East direction will be

Total displacement towards South

now to find the net displacement we can use vector addition



<em>so magnitude of net displacement will be equal to 13 blocks</em>
To solve this problem, we must imagine that Jim’s initial
position, the position of the rock, and Jim’s final position all connects to
form a triangle. Now we can imagine that the triangle is a right triangle with
the 90° angle on the initial position.
The angle of 30° is directly opposite to the length of his
total stride while the width of the river is the side adjacent to the angle.
Therefore can use the tan function to solve for the width of the river:
tan θ = opposite side / adjacent side
tan 30 = total stride distance / width of river
where total stride distance = 65 * 0.8 = 52 m
width of river = 52 m / tan 30
<span>width of river = 90.07 m</span>
<span>Acceleration is the change in velocity divided by time taken. It has both magnitude and direction. In this problem, the change in velocity would first have to be calculated. Velocity is distance divided by time. Therefore, the velocity here would be 300 m divided by 22.4 seconds. This gives a velocity of 13.3928 m/s. Since acceleration is velocity divided by time, it would be 13.3928 divided by 22.4, giving a final solution of 0.598 m/s^2.</span>