Question

A particle is located at the origin. A second particle is a distance D away in a direction [E A N]. At the same time, the first and second particles suddenly move at constant speeds |v1| and |v2| in directions [E Θ1 N] and [W Θ2 N] respectively. If D = 25m, A = 30°, Θ1 = 50°, and Θ2 = 35°, and |v1| = 5.3m/s, what is |v2|, and at what time do they meet?

Answer 1

To find the speed of the second particle (v2), we can use the law of cosines. The law of cosines states that for a triangle with sides a, b, and c, and the angle opposite side c denoted as θ, the following equation holds true: c^2 = a^2 + b^2 - 2ab*cos(θ). In this case, we know that side a (|v1|) has a magnitude of 5.3 m/s, side b (|v2|) is unknown, and side c (D) has a magnitude of 25 m. The angle θ is the difference between Θ1 and Θ2, so θ = 50° - 35° = 15°. Plugging in the values into the law of cosines, we have 25^2 = 5.3^2 + |v2|^2 - 2*5.3*|v2|*cos(15°). Solving this equation gives us |v2| ≈ 6.64 m/s. To find the time at which they meet, we can use the formula t = D / (|v1| + |v2|). Substituting in the values, we have t = 25 / (5.3 + 6.64) ≈ 1.92 s.

Step-by-step explanation:

To find the speed of the second particle (v2), we can use the law of cosines. The law of cosines states that for a triangle with sides a, b, and c, and the angle opposite side c denoted as θ, the following equation holds true: c^2 = a^2 + b^2 - 2ab*cos(θ).

In this case, we know that side a (|v1|) has a magnitude of 5.3 m/s, side b (|v2|) is unknown, and side c (D) has a magnitude of 25 m. The angle θ is the difference between Θ1 and Θ2, so θ = 50° - 35° = 15°.

Plugging in the values into the law of cosines, we have 25^2 = 5.3^2 + |v2|^2 - 2*5.3*|v2|*cos(15°). Solving this equation gives us |v2| ≈ 6.64 m/s.

To find the time at which they meet, we can use the formula t = D / (|v1| + |v2|). Substituting in the values, we have t = 25 / (5.3 + 6.64) ≈ 1.92 s.