Question

A particle is constrained to move in the xy-plane. Its velocity is described by the vector function: $\vec{v}(t) = (5 \cdot \sin(\pi t/2) \cdot \hat{i} + 6e^{-t/5} \cdot \hat{j}) \text{ m/s}$ for $t \ge 0 \text{ s}$. If its position at $t = 0 \text{ s}$ is: $\vec{r}(0) = (-2.3 \cdot \hat{i} + 4.8 \cdot \hat{j}) \text{ m}$, then what is the direction of the particle's position vector (in degrees, in standard form) at time $t = 2.5 \text{ s}$?

Name: A particle is constrained to move in the xy-plane. Its velocity is described by the vector function: v⃗(t) = (5 ·sin(π t/2) ·î + 6e^-t/5·ĵ) m/s for t ≥ 0 s. If its position at t = 0 s is: r⃗(0) = (-2.3 ·î + 4.8 ·ĵ) m, then what is the direction of the particle's position vector (in degrees, in standard form) at time t = 2.5 s?
Uploaded: 2023-06-19T02:01:54-08:00
Duration: 1 min 45 s
Channel: Adi S
Description: A particle is constrained to move in the xy-plane. Its velocity is described by the vector function: v⃗(t) = (5 ·sin(π t/2) ·î + 6e^-t/5·ĵ) m/s for t ≥ 0 s. If its position at t = 0 s is: r⃗(0) = (-2.3 ·î + 4.8 ·ĵ) m, then what is the direction of the particle's position vector (in degrees, in standard form) at time t = 2.5 s?

          A particle is constrained to move in the xy-plane. Its velocity is described by the vector function: $\vec{v}(t) = (5 \cdot \sin(\pi t/2) \cdot \hat{i} + 6e^{-t/5} \cdot \hat{j}) \text{ m/s}$ for $t \ge 0 \text{ s}$. If its position at $t = 0 \text{ s}$ is: $\vec{r}(0) = (-2.3 \cdot \hat{i} + 4.8 \cdot \hat{j}) \text{ m}$, then what is the direction of the particle's position vector (in degrees, in standard form) at time $t = 2.5 \text{ s}$?

A particle is constrained to move in the xy-plane. Its velocity is described by the vector function: v⃗(t) = (5 ·sin(π t/2) ·î + 6e^-t/5·ĵ) m/s for t ≥ 0 s. If its position at t = 0 s is: r⃗(0) = (-2.3 ·î + 4.8 ·ĵ) m, then what is the direction of the particle's position vector (in degrees, in standard form) at time t = 2.5 s?

Added by Peggy B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

06/19/2023

Step 1

To find r(t), we need to integrate v(t) with respect to t. r(t) = ∫v(t) dt Given that v(t) = 5sin(rt/2 + 6e^(-t/5).j m/s, we can integrate each component separately. The x-component of r(t) is given by: r_x(t) = ∫(5sin(rt/2 + 6e^(-t/5))) dt To integrate Show more…

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A particle is constrained to move in the xy-plane. Its velocity is described by the vector function v(t) = 5sin(rt/2 + 6e^(-t/5).j m/s for t ≥ 0 s. If its position at t = 0 s is r(0) = (-2.3 + 4.8j) m, then what is the direction of the particle's position vector (in degrees, in standard form) at time t = 2.5 s?