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.

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;

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

The radial acceleration of the rocket is calculated as follows;

The tangential acceleration is calculated as follows;

The resultant acceleration of the rocket is calculated as follows;

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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

the focus of lesson 1 is newton's first law of motion - sometimes referred to as the law of inertia. newton's first law of motion is often stated as. an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

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