Question

The resultant of the velocity of 3 m/s at 0° and 4 m/s at 60° is​

Answer 1

Explanation:

apply the formula for resultant vector and get the answer

Answer 2

Answer:

The resultant of the velocity of 3 m/s at 0° and 4 m/s at 60° is √37 m/sec .

Explanation:

Given that :

• Magnitude of first vector, a = 3 m/sec
• Magnitude of second vector, b = 4 m/sec
• Angle between two vectors, θ = 60⁰

Solution :

• The resultant of two vectors whose magnitudes are a and b and θ is the angle between them, is given by:

$R= \sqrt{ {a}^{2} + {b}^{2} + 2ab \cos θ }$

• The resultant of vectors having magnitude 3 m/sec and 4 m/sec and having 60⁰ angle between them is

$R = \sqrt{ {3}^{2} + {4}^{2} +(2 \times 3 \times 4 \times ) \cos60⁰} \\ = \sqrt{9 + 16 + 12} \\ = \sqrt{37}$

• Hence, resultant of velocity 3 m/sec and 4 m/sec having 60⁰ angle between them is √37 m/sec .