Question
1. Determine the displacement vector that must be added to the displacement (23i-16j)m ti give displacement of 7.0m pointing in the +x-direction
2. A force (15i-16j+27k)N is added to a force of (23j-40k)N what is the magnitude of the resultant?
3. A truck is moving north at a speed of 70km/h. The exhaust pipe above the truck cap sends out a trial of smoke that makes 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?
4. A shop is traveling due east at 10km/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
5. A boat, propelled so as to travel with a speed of 0.50m/s in still water, moves directly across a river that is 60m wide. The river flow with a speed of 0.30m/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 ?
6. A reckless drunk is playing with a gun in an airplane that is going directly east at 500km/h. The drunk shoots the gun straight up at the ceiling of the plane. The bullet leaves the the ground at a speed of 1000km/h. According to someone standing on the Earth what angle does the bullet make with the vertical?
2. A force (15i-16j+27k)N is added to a force of (23j-40k)N what is the magnitude of the resultant?
3. A truck is moving north at a speed of 70km/h. The exhaust pipe above the truck cap sends out a trial of smoke that makes 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?
4. A shop is traveling due east at 10km/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
5. A boat, propelled so as to travel with a speed of 0.50m/s in still water, moves directly across a river that is 60m wide. The river flow with a speed of 0.30m/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 ?
6. A reckless drunk is playing with a gun in an airplane that is going directly east at 500km/h. The drunk shoots the gun straight up at the ceiling of the plane. The bullet leaves the the ground at a speed of 1000km/h. According to someone standing on the Earth what angle does the bullet make with the vertical?
Answers
GPT-4o mini
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:
1. An upward component \( 1000 \, \text{km/h} \) (vertical).
2. 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:
\[
\boxed{26.57^\circ}
\]
### 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:
1. An upward component \( 1000 \, \text{km/h} \) (vertical).
2. 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:
\[
\boxed{26.57^\circ}
\]