Question

The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give the values r = 32700 ft, r¨ = 85 ft/sec2, and θ˙ = 0.019 rad/sec. Calculate the magnitudes of the velocity and acceleration of the rocket at this position.

Answer 1

Answer:

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

a = (85 - 11.8047)/0.9135

a = 80.13 ft/s²

Answer 2

The magnitude of the linear velocity of the rocket at the given position is 1,667.2 ft/s.

The magnitude of the acceleration of the rocket at the given position is 90.73 ft/s².

The given parameters;

• angular distance, θ = 66⁰
• radius of the path, r = 32700 ft
• centripetal acceleration, ac = 85 ft/s²
• initial angular velocity, ωi = 0.019 rad/s

The linear velocity and final angular velocity of the rocket is calculated as follows;

$a_c = \frac{v^2}{r} \\\\v^2 = a_c \times r\\\\v = \sqrt{a_c r} \\\\v = \sqrt{85 \times 32700}\\\\v = 1667.2 \ ft/s\\\\\omega_f = \frac{v}{r} = \frac{1667.2}{32700} = 0.051 \ rad/s$

The radial acceleration of the rocket is calculated as follows;

$\omega_f^2 = \omega_i^2 + 2\alpha _r \theta\\\\(0.051)^2 = (0.019)^2 + 2(66^0 \times \frac{2\pi \ rad}{360 ^0} )(\alpha _r)\\\\(0.051)^2 - (0.019)^2 = 2(66^0 \times \frac{2\pi \ rad}{360 ^0} )(\alpha _r) \\\\0.00224 = 2.304\alpha _r\\\\\alpha _r = \frac{0.00224 }{ 2.304} \\\\\alpha _r = 0.00097 \ rad/s^2$

The tangential acceleration is calculated as follows;

$a_t = \alpha_r \times r\\\\a_t = 0.00097 \times 32700\\\\a_t = 31.72 \ ft/s^2$

The resultant acceleration of the rocket is calculated as follows;

$a = \sqrt{a_t^2 + a_c^2} \\\\a = \sqrt{31.72^2 \ + \ 85^2} \\\\a= 90.73 \ ft/s^2$

Learn more here:https://brainly.com/question/14086497