Answer :
Certainly! Let's address each part of the question with a detailed, step-by-step solution.
### (i) Vector Diagram
To visualize this, you should sketch the scenario:
- Draw a horizontal line representing the ground.
- Draw another horizontal line above the ground line; this represents the flight path of the aeroplane.
- Draw an inclined line starting from a point on the ground (representing where the missile is fired from) at an angle of [tex]\(35^\circ\)[/tex] to the horizontal; this line represents the missile's trajectory.
- Label the following:
- The angle between the missile's trajectory and the horizontal ground as [tex]\(35^\circ\)[/tex].
- The horizontal velocity vector of the aeroplane as [tex]\(v = 150 \, \text{m/s}\)[/tex].
- The missile's acceleration along its trajectory as [tex]\(a = 50 \, \text{m/s}^2\)[/tex].
### (ii) Show that the time taken by the missile to reach the aeroplane is [tex]\( t = \frac{2v}{a \cos 35} \)[/tex]
To derive this formula, follow these steps:
1. Break Down the Acceleration Component:
The missile is accelerating with a magnitude of [tex]\(a\)[/tex] at an angle of [tex]\(35^\circ\)[/tex] to the horizontal. The horizontal component of this acceleration is:
[tex]\[ a_x = a \cos 35^\circ \][/tex]
2. Missile's Horizontal Motion:
Since the missile and the aeroplane must come into alignment along their trajectories for the missile to hit the aeroplane:
[tex]\[ \text{Horizontal distance} = \text{Horizontal velocity of the aeroplane} \times \text{time} \][/tex]
Since the aeroplane travels horizontally at a constant speed [tex]\(v\)[/tex], the distance the aeroplane travels in time [tex]\(t\)[/tex] is [tex]\( vt \)[/tex].
3. Equating the Distances:
The horizontal distance covered by the missile, given its uniform horizontal acceleration, can be expressed using the equation:
[tex]\[ \text{Horizontal distance} = \frac{1}{2} a_x t^2 = \frac{1}{2} (a \cos 35^\circ) t^2 \][/tex]
4. Set the Distances Equal:
For the missile to hit the aeroplane:
[tex]\[ vt = \frac{1}{2} (a \cos 35^\circ) t^2 \][/tex]
5. Solve for [tex]\(t\)[/tex]:
Dividing both sides by [tex]\(t\)[/tex]:
[tex]\[ v = \frac{1}{2} a \cos 35^\circ \cdot t \][/tex]
Rearranging for [tex]\(t\)[/tex]:
[tex]\[ t = \frac{2v}{a \cos 35^\circ} \][/tex]
This derivation confirms that the time [tex]\(t\)[/tex] is given by:
[tex]\[ t = \frac{2v}{a \cos 35^\circ} \][/tex]
### (iii) Calculate the time
Now, let's use the given values to calculate [tex]\( t \)[/tex]:
- Given:
[tex]\[ v = 150 \, \text{m/s}, \quad a = 50 \, \text{m/s}^2, \quad \text{and} \quad \cos 35^\circ \approx 0.8192 \][/tex]
- Substitute these values into the formula:
[tex]\[ t = \frac{2 \times 150}{50 \times 0.8192} \][/tex]
- After performing these substitutions and calculations, we find:
[tex]\[ t \approx 7.32 \, \text{seconds} \][/tex]
So, the time taken by the missile to reach the aeroplane is approximately [tex]\( 7.32 \)[/tex] seconds.
### (i) Vector Diagram
To visualize this, you should sketch the scenario:
- Draw a horizontal line representing the ground.
- Draw another horizontal line above the ground line; this represents the flight path of the aeroplane.
- Draw an inclined line starting from a point on the ground (representing where the missile is fired from) at an angle of [tex]\(35^\circ\)[/tex] to the horizontal; this line represents the missile's trajectory.
- Label the following:
- The angle between the missile's trajectory and the horizontal ground as [tex]\(35^\circ\)[/tex].
- The horizontal velocity vector of the aeroplane as [tex]\(v = 150 \, \text{m/s}\)[/tex].
- The missile's acceleration along its trajectory as [tex]\(a = 50 \, \text{m/s}^2\)[/tex].
### (ii) Show that the time taken by the missile to reach the aeroplane is [tex]\( t = \frac{2v}{a \cos 35} \)[/tex]
To derive this formula, follow these steps:
1. Break Down the Acceleration Component:
The missile is accelerating with a magnitude of [tex]\(a\)[/tex] at an angle of [tex]\(35^\circ\)[/tex] to the horizontal. The horizontal component of this acceleration is:
[tex]\[ a_x = a \cos 35^\circ \][/tex]
2. Missile's Horizontal Motion:
Since the missile and the aeroplane must come into alignment along their trajectories for the missile to hit the aeroplane:
[tex]\[ \text{Horizontal distance} = \text{Horizontal velocity of the aeroplane} \times \text{time} \][/tex]
Since the aeroplane travels horizontally at a constant speed [tex]\(v\)[/tex], the distance the aeroplane travels in time [tex]\(t\)[/tex] is [tex]\( vt \)[/tex].
3. Equating the Distances:
The horizontal distance covered by the missile, given its uniform horizontal acceleration, can be expressed using the equation:
[tex]\[ \text{Horizontal distance} = \frac{1}{2} a_x t^2 = \frac{1}{2} (a \cos 35^\circ) t^2 \][/tex]
4. Set the Distances Equal:
For the missile to hit the aeroplane:
[tex]\[ vt = \frac{1}{2} (a \cos 35^\circ) t^2 \][/tex]
5. Solve for [tex]\(t\)[/tex]:
Dividing both sides by [tex]\(t\)[/tex]:
[tex]\[ v = \frac{1}{2} a \cos 35^\circ \cdot t \][/tex]
Rearranging for [tex]\(t\)[/tex]:
[tex]\[ t = \frac{2v}{a \cos 35^\circ} \][/tex]
This derivation confirms that the time [tex]\(t\)[/tex] is given by:
[tex]\[ t = \frac{2v}{a \cos 35^\circ} \][/tex]
### (iii) Calculate the time
Now, let's use the given values to calculate [tex]\( t \)[/tex]:
- Given:
[tex]\[ v = 150 \, \text{m/s}, \quad a = 50 \, \text{m/s}^2, \quad \text{and} \quad \cos 35^\circ \approx 0.8192 \][/tex]
- Substitute these values into the formula:
[tex]\[ t = \frac{2 \times 150}{50 \times 0.8192} \][/tex]
- After performing these substitutions and calculations, we find:
[tex]\[ t \approx 7.32 \, \text{seconds} \][/tex]
So, the time taken by the missile to reach the aeroplane is approximately [tex]\( 7.32 \)[/tex] seconds.
