Let’s break down each problem one by one.
Problem 1:
Determine the displacement vector that must be added to the displacement \( \mathbf{d} = 23\mathbf{i} - 16\mathbf{j} , \text{m} \) to give a displacement of \( 7.0 , \text{m} \) pointing in the \( +x \)-direction.
The displacement vector pointing in the \( +x \)-direction with a magnitude of \( 7.0 , \text{m} \) can be represented as: \[ \mathbf{d}{\text{target}} = 7\mathbf{i} \] The vector that must be added to \( \mathbf{d} \) is: \[ \mathbf{d}{\text{add}} = \mathbf{d}{\text{target}} - \mathbf{d} \] Substituting the values: \[ \mathbf{d}{\text{add}} = 7\mathbf{i} - (23\mathbf{i} - 16\mathbf{j}) = (7 - 23)\mathbf{i} + 16\mathbf{j} = -16\mathbf{i} + 16\mathbf{j} \] Thus, the displacement vector that must be added is: \[ \boxed{-16\mathbf{i} + 16\mathbf{j} , \text{m}} \]
Problem 2:
A force \( \mathbf{F}_1 = 15\mathbf{i} - 16\mathbf{j} + 27\mathbf{k} , \text{N} \) is added to a force \( \mathbf{F}_2 = 23\mathbf{j} - 40\mathbf{k} , \text{N} \). What is the magnitude of the resultant?
First, calculate the resultant force \( \mathbf{F}{\text{R}} \): \[ \mathbf{F}{\text{R}} = \mathbf{F}1 + \mathbf{F}2 = (15\mathbf{i} - 16\mathbf{j} + 27\mathbf{k}) + (0\mathbf{i} + 23\mathbf{j} - 40\mathbf{k}) \] \[ = 15\mathbf{i} + (-16 + 23)\mathbf{j} + (27 - 40)\mathbf{k} = 15\mathbf{i} + 7\mathbf{j} - 13\mathbf{k} \] Now, compute the magnitude \( |\mathbf{F}{\text{R}}| \): \[ |\mathbf{F}{\text{R}}| = \sqrt{15^2 + 7^2 + (-13)^2} = \sqrt{225 + 49 + 169} = \sqrt{443} \] Thus, the magnitude of the resultant is: \[ \boxed{\sqrt{443} , \text{N}} \quad \approx 21.0 , \text{N} \]
Problem 3:
A truck is moving north at a speed of 70 km/h. The exhaust pipe above the truck cap sends out a trail of smoke that makes an angle of 20° east of south behind the truck. If the wind is blowing directly towards the east, what is the wind speed at that location?
To find the wind speed, we can represent the velocities as vectors. The truck's velocity vector \( \mathbf{V}{\text{truck}} \) is: \[ \mathbf{V}{\text{truck}} = 70\mathbf{j} , \text{km/h} \] The smoke makes an angle of \( 20° \) east of south, hence the smoke trail's velocity relative to the ground can be expressed as: \[ \mathbf{V}{\text{smoke}} = V{s} (\sin(20°)\mathbf{i} - \cos(20°)\mathbf{j}) \] Since the smoke's \( y \)-component must compensate for the truck's velocity: \[ V_{s} \cos(20°) = 70 \] Solving for \( V_s \): \[ V_s = \frac{70}{\cos(20°)} \approx \frac{70}{0.9397} \approx 74.5 , \text{km/h} \] Now, compute the eastward component due to wind: \[ V_{\text{east}} = V_s \sin(20°) \approx 74.5 \times 0.3420 \approx 25.5 , \text{km/h} \] Therefore, the wind speed at that location is: \[ \boxed{25.5 , \text{km/h}} \]
Problem 4:
A shop is traveling due east at 10 km/h. What must be the speed of a second ship heading 30° east of North if it is always due north of the first ship?
Let \( v_{2} \) be the speed of the second ship \( S_2 \). The eastward component of \( S_2 \) can be expressed as: \[ v_{2, x} = v_{2} \sin(30°) = \frac{v_{2}}{2} \] For \( S_2 \) to remain due north of \( S_1 \): \[ v_{2, x} = 10 \rightarrow \frac{v_{2}}{2} = 10 \rightarrow v_{2} = 20 , \text{km/h} \] The required speed of the second ship is: \[ \boxed{20 , \text{km/h}} \]
Problem 5:
A boat, propelled so as to travel with a speed of 0.50 m/s in still water, moves directly across a river that is 60 m wide. The river flows with a speed of 0.30 m/s. (a) At what angle relative to the straight-across direction must the boat be pointed? (b) How long does it take the boat to cross the river?
(a) Let’s denote:
- Velocity of the boat across the river = \( V_b = 0.50 , \text{m/s} \)
- Velocity of the river = \( V_r = 0.30 , \text{m/s} \)
Let \( \theta \) be the angle at which the boat is aimed relative to the riverbank.
To calculate the angle \( \theta \), we can use the following relation: \[ \tan(\theta) = \frac{V_r}{V_b} \] Therefore: \[ \theta = \tan^{-1}( \frac{0.30}{0.50}) \approx \tan^{-1}(0.6) \approx 30.96^\circ \] Thus, the angle relative to the straight-across direction must be: \[ \boxed{30.96^\circ , \text{upstream}} \]
(b) The time taken to cross the river can be calculated using: \[ t = \frac{\text{width of river}}{\text{velocity across river}} = \frac{60 , \text{m}}{0.50 , \text{m/s}} = 120 , \text{s} \] Therefore, the time taken to cross the river is: \[ \boxed{120 , \text{s}} \]
Problem 6:
A reckless drunk is playing with a gun in an airplane that is going directly east at 500 km/h. The drunk shoots the gun straight up at the ceiling of the plane. The bullet leaves the ground at a speed of 1000 km/h. According to someone standing on the Earth, what angle does the bullet make with the vertical?
The bullet has two velocity components when it is shot straight up:
- An upward component \( 1000 , \text{km/h} \) (vertical).
- A horizontal component – since the airplane is moving eastward – it also has a horizontal velocity \( 500 , \text{km/h} \).
To find the angle \( \theta \) with respect to the vertical, we can use the tangent function: \[ \tan(\theta) = \frac{\text{horizontal velocity}}{\text{vertical velocity}} = \frac{500}{1000} = 0.5 \] Thus, we find: \[ \theta = \tan^{-1}(0.5) \approx 26.57^\circ \] Therefore, the angle that the bullet makes with the vertical is: \[ \boxed{26.57^\circ} \]