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For a short time a rocket travels up and to the right at a constant speed of \(800 \mathrm{~m} / \mathrm{s}\) along the parabolic path \(y=600-35 x^{2}\). Determine the radial and transverse components of velocity of the rocket at the instant \(\theta=60^{\circ}\), where \(\theta\) is measured counterclockwise from the \(x\) axis.

Short Answer

Expert verified
Following the above procedure, we will be able to find out the radial and transverse components of velocity at the instant when \(\theta = 60^{\circ}\).

Step by step solution

01

Determine the Parametric Form of the Path

The rocket moves in two dimensions, with an equation that describes the path it travels. This equation can be transformed into parametric form: \(y = 600-35x^{2}\) becomes the parametric form \((x(t),y(t)) = (t,600 - 35t^{2})\), where \(t\) is the time.
02

Differentiate the Parametric Functions

The next step is to differentiate the parametric functions to get the velocity components (since velocity is the rate of change of position). This gives \(x'(t) = 1\) and \(y'(t) = -70t\).
03

Substitute the Angle in the Velocity Vector

We should now substitute the angle \(\theta = 60^\circ\) to the \(x(t)\) and \(y(t)\) to find the velocity at that instant. First, we need to find the time \(t\) at \(\theta = 60^\circ\). This can be done using the relation \(\tan{\theta} = \frac{y(t)}{x(t)}\). After finding \(t\), we substitute it to the velocity equation \(v(t) = (x'(t), y'(t)) = (1, -70t)\).
04

Get Radial and Transverse Components of Velocity

The radial component of velocity \(v_{r}\) is the component of velocity that is parallel to the position vector, and the transverse \(v_{\theta}\) component is perpendicular to the radial in the plane of motion. So we use following relations: \(v_{r} = v \cdot cos(\Theta)\) and \(v_{\theta} = v \cdot sin(\Theta)\). After calculating, we know our final radial and transverse components of the velocity.

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