Answer :
Certainly! Let's solve this problem step by step.
1. Given data:
- We have two equal velocities.
- The resultant of these velocities is [tex]\(1 \frac{1}{2}\)[/tex] times the value of either velocity.
2. Interpretation:
- Assume each velocity is [tex]\( v \)[/tex].
- The resultant velocity, [tex]\( R \)[/tex], is [tex]\(1 \frac{1}{2}v = \frac{3}{2}v = 1.5v\)[/tex].
3. Formula for the resultant of two vectors:
The formula for the magnitude of the resultant vector, [tex]\( R \)[/tex], when two vectors of magnitudes [tex]\( v_1 \)[/tex] and [tex]\( v_2 \)[/tex] are combined at an angle [tex]\( \theta \)[/tex] between them, is:
[tex]\[ R = \sqrt{v_1^2 + v_2^2 + 2v_1v_2 \cos(\theta)} \][/tex]
Since we have equal magnitudes for both vectors, [tex]\( v_1 = v_2 = v \)[/tex]. So, the formula simplifies to:
[tex]\[ R = \sqrt{v^2 + v^2 + 2v^2 \cos(\theta)} = \sqrt{2v^2 + 2v^2 \cos(\theta)} \][/tex]
[tex]\[ R = v \sqrt{2 + 2 \cos(\theta)} \][/tex]
4. Given relation between the resultant and the individual velocity:
According to the problem, the resultant [tex]\( R \)[/tex] is [tex]\(1.5v\)[/tex]. Thus:
[tex]\[ 1.5v = v \sqrt{2 + 2 \cos(\theta)} \][/tex]
Dividing both sides by [tex]\( v \)[/tex], we get:
[tex]\[ 1.5 = \sqrt{2 + 2 \cos(\theta)} \][/tex]
5. Solving for [tex]\( \cos(\theta) \)[/tex]:
Square both sides to eliminate the square root:
[tex]\[ 1.5^2 = 2 + 2 \cos(\theta) \][/tex]
[tex]\( 1.5^2 = 2.25 \)[/tex], so:
[tex]\[ 2.25 = 2 + 2 \cos(\theta) \][/tex]
Solving for [tex]\( \cos(\theta) \)[/tex]:
[tex]\[ 2.25 - 2 = 2 \cos(\theta) \][/tex]
[tex]\[ 0.25 = 2 \cos(\theta) \][/tex]
[tex]\[ \cos(\theta) = \frac{0.25}{2} = 0.125 \][/tex]
6. Finding the angle [tex]\( \theta \)[/tex]:
To find the angle [tex]\( \theta \)[/tex], take the inverse cosine (arccos) of [tex]\( 0.125 \)[/tex]:
[tex]\[ \theta = \arccos(0.125) \][/tex]
Using a calculator or table to find the arccos value:
[tex]\[ \theta \approx 1.445 \text{ radians} \][/tex]
Convert this to degrees:
[tex]\[ \theta \approx 82.82^\circ \][/tex]
Therefore, we conclude that the angle between the two velocities is approximately [tex]\( 82.82^\circ \)[/tex].
### Verification of the Other Answer:
The problem mentions another possible angle answer of [tex]\( 60^\circ \)[/tex]. To verify if this might be another solution, we would have to see if [tex]\( \cos(\theta) \)[/tex] of [tex]\( 60^\circ \)[/tex] (which is [tex]\( 0.5 \)[/tex]) satisfies the given conditions. Given our calculation, [tex]\( \cos(\theta) \approx 0.125 \)[/tex] is the unique solution based on the resultant being [tex]\( 1.5v \)[/tex].
Thus, the primary and accurate answer relevant to the stated problem is:
[tex]\[ \boxed{82.82^\circ} \][/tex]
1. Given data:
- We have two equal velocities.
- The resultant of these velocities is [tex]\(1 \frac{1}{2}\)[/tex] times the value of either velocity.
2. Interpretation:
- Assume each velocity is [tex]\( v \)[/tex].
- The resultant velocity, [tex]\( R \)[/tex], is [tex]\(1 \frac{1}{2}v = \frac{3}{2}v = 1.5v\)[/tex].
3. Formula for the resultant of two vectors:
The formula for the magnitude of the resultant vector, [tex]\( R \)[/tex], when two vectors of magnitudes [tex]\( v_1 \)[/tex] and [tex]\( v_2 \)[/tex] are combined at an angle [tex]\( \theta \)[/tex] between them, is:
[tex]\[ R = \sqrt{v_1^2 + v_2^2 + 2v_1v_2 \cos(\theta)} \][/tex]
Since we have equal magnitudes for both vectors, [tex]\( v_1 = v_2 = v \)[/tex]. So, the formula simplifies to:
[tex]\[ R = \sqrt{v^2 + v^2 + 2v^2 \cos(\theta)} = \sqrt{2v^2 + 2v^2 \cos(\theta)} \][/tex]
[tex]\[ R = v \sqrt{2 + 2 \cos(\theta)} \][/tex]
4. Given relation between the resultant and the individual velocity:
According to the problem, the resultant [tex]\( R \)[/tex] is [tex]\(1.5v\)[/tex]. Thus:
[tex]\[ 1.5v = v \sqrt{2 + 2 \cos(\theta)} \][/tex]
Dividing both sides by [tex]\( v \)[/tex], we get:
[tex]\[ 1.5 = \sqrt{2 + 2 \cos(\theta)} \][/tex]
5. Solving for [tex]\( \cos(\theta) \)[/tex]:
Square both sides to eliminate the square root:
[tex]\[ 1.5^2 = 2 + 2 \cos(\theta) \][/tex]
[tex]\( 1.5^2 = 2.25 \)[/tex], so:
[tex]\[ 2.25 = 2 + 2 \cos(\theta) \][/tex]
Solving for [tex]\( \cos(\theta) \)[/tex]:
[tex]\[ 2.25 - 2 = 2 \cos(\theta) \][/tex]
[tex]\[ 0.25 = 2 \cos(\theta) \][/tex]
[tex]\[ \cos(\theta) = \frac{0.25}{2} = 0.125 \][/tex]
6. Finding the angle [tex]\( \theta \)[/tex]:
To find the angle [tex]\( \theta \)[/tex], take the inverse cosine (arccos) of [tex]\( 0.125 \)[/tex]:
[tex]\[ \theta = \arccos(0.125) \][/tex]
Using a calculator or table to find the arccos value:
[tex]\[ \theta \approx 1.445 \text{ radians} \][/tex]
Convert this to degrees:
[tex]\[ \theta \approx 82.82^\circ \][/tex]
Therefore, we conclude that the angle between the two velocities is approximately [tex]\( 82.82^\circ \)[/tex].
### Verification of the Other Answer:
The problem mentions another possible angle answer of [tex]\( 60^\circ \)[/tex]. To verify if this might be another solution, we would have to see if [tex]\( \cos(\theta) \)[/tex] of [tex]\( 60^\circ \)[/tex] (which is [tex]\( 0.5 \)[/tex]) satisfies the given conditions. Given our calculation, [tex]\( \cos(\theta) \approx 0.125 \)[/tex] is the unique solution based on the resultant being [tex]\( 1.5v \)[/tex].
Thus, the primary and accurate answer relevant to the stated problem is:
[tex]\[ \boxed{82.82^\circ} \][/tex]
