Question

A particle P starts at time t = 0 s at the position $$x = 10 m$$ $$y = -13 m.$$ The velocity $\vec{v}$ of the particle is written in the polar basis associated with its current position, and is: $$\vec{v} = \hat{e}_r + \hat{e}_\theta m/s.$$ What is the position of P at t = 3 s?

Name: a particle p starts at time t 0 s at the position x 10 m y 13 m the velocity vecv of the particle is written in the polar basis associated with its current position and is vecv hater hatet 36085
Uploaded: 2023-12-13T05:26:49-08:00
Duration: 6 min 34 s
Channel: Adi S
Description: a particle p starts at time t 0 s at the position x 10 m y 13 m the velocity vecv of the particle is written in the polar basis associated with its current position and is vecv hater hatet 36085

          A particle P starts at time t = 0 s at the position
$$x = 10 m$$
$$y = -13 m.$$
The velocity $\vec{v}$ of the particle is written in the polar basis associated
with its current position, and is:
$$\vec{v} = \hat{e}_r + \hat{e}_\theta m/s.$$
What is the position of P at t = 3 s?

a particle p starts at time t 0 s at the position x 10 m y 13 m the velocity vecv of the particle is written in the polar basis associated with its current position and is vecv hater hatet 36085

Added by Gina S.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

12/13/2023

Step 1

$$r = \sqrt{x^2 + y^2} = \sqrt{10^2 + (-13)^2} = \sqrt{100 + 169} = \sqrt{269} \approx 16.4 m$$ $$\theta = \arctan(\frac{y}{x}) = \arctan(\frac{-13}{10}) \approx -0.92 rad$$ Show more…

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A particle P starts at time t = 0 s at the position $$x = 10 m$$ $$y = -13 m.$$ The velocity $\vec{v}$ of the particle is written in the polar basis associated with its current position, and is: $$\vec{v} = \hat{e}_r + \hat{e}_\theta m/s.$$ What is the position of P at t = 3 s?