Question

Math

Posted 7 months ago

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7. At time $t \geq 0$, a particle moving in the $x y$-plane has velocity vector given by $v(t)=\left\langle 4 e^{-t}, \sin (1+\sqrt{t})\right\rangle$. What is the total distance the particle travels between $t=1$ and $t=3$ ?
(A) 1.861
(B) 1.983
(C) 2.236
(D) 4.851
```

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Answer from Sia

Posted 7 months ago

Solution by Steps

step 1

To find the total distance the particle travels, we need to integrate the magnitude of the velocity vector over the given time interval

step 2

The magnitude of the velocity vector $v(t)$ is given by $\sqrt{(4e^{-t})^2 + (\sin(1+\sqrt{t}))^2}$

step 3

We integrate this magnitude from $t=1$ to $t=3$ to find the total distance traveled

step 4

The asksia-ll calculator has already computed this integral: $\int_{1}^{3} \sqrt{16 e^{-2t} + \sin^2(1 + \sqrt{t})} \, dt = 1.8606$

Answer

(A) 1.861

Key Concept

Total Distance Traveled by a Particle

Explanation

The total distance traveled by a particle is the integral of the magnitude of its velocity vector over the time interval.

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