A particle moving in the $x y$-plane has velocity vector given by
$v(t)=(-12,-5 t)$ for time $\boldsymbol{t} \geq 0$. At $\boldsymbol{t}=3$, the particle is at the point $(-1,-2)$.

What is the particle's position at $t=5$ ?

Question

A particle moving in the $x y$-plane has velocity vector given by
$v(t)=(-12,-5 t)$ for time $\boldsymbol{t} \geq 0$. At $\boldsymbol{t}=3$, the particle is at the point $(-1,-2)$.

What is the particle's position at $t=5$ ?

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the particle's position at $t=5$, we first need to integrate the velocity vector components separately. The first component of the velocity vector is $v_x(t) = -12$. ‖ ‖ step 2 ⋮ Integrate $v_x(t)$ from $t=3$ to $t=5$ to find the change in the $x$-position. $\int_3^5 -12 \, dt = -24$. ‖ ‖ step 3 ⋮ The initial $x$-position at $t=3$ is $x(3) = -1$. Add the change in $x$-position to this to find $x(5)$. $x(5) = x(3) + (-24) = -1 - 24 = -25$. ‖ ‖ step 4 ⋮ The second component of the velocity vector is $v_y(t) = -5t$. ‖ ‖ step 5 ⋮ Integrate $v_y(t)$ from $t=3$ to $t=5$ to find the change in the $y$-position. $\int_3^5 -5t \, dt = -40$. ‖ ‖ step 6 ⋮ The initial $y$-position at $t=3$ is $y(3) = -2$. Add the change in $y$-position to this to find $y(5)$. $y(5) = y(3) + (-40) = -2 - 40 = -42$. ‖ ‖ step 7 ⋮ Combine the $x$ and $y$ positions to find the particle's position at $t=5$. The position is given by the point $(x(5), y(5))$. ‖ ∻Answer∻ ⚹ The particle's position at $t=5$ is $(-25, -42)$. ⚹ ∻Key Concept∻ ⚹Integration of velocity to find displacement⚹ ∻Explanation∻ ⚹The particle's position at a given time is found by integrating the velocity vector components over the time interval and adding the result to the initial position.⚹◊What is the particle's position at $t=5$ in the $x y$-plane?⍭ Generate me a similar question◊

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