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# Problem: A particle's trajectory is described by x =(12t3−2t2)m and y =(12t2−2t)m, where t is in s.Part A What is the particle's speed at t=0s ? v = m/s

###### FREE Expert Solution

Velocity:

$\overline{){\mathbf{v}}{\mathbf{=}}\sqrt{{{\mathbf{v}}_{\mathbf{x}}}^{\mathbf{2}}\mathbf{+}{{\mathbf{v}}_{\mathbf{y}}}^{\mathbf{2}}}}$

The x-component of velocity, vx

$\begin{array}{rcl}{\mathbf{v}}_{\mathbf{x}}& \mathbf{=}& \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\\ & \mathbf{=}& \frac{\mathbf{d}}{\mathbf{d}\mathbf{t}}\mathbf{\left(}\mathbf{12}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{2}{\mathbf{t}}^{\mathbf{2}}\mathbf{\right)}\end{array}$

vx = 36t2 - 4t

The y-component of velocity, vy

$\begin{array}{rcl}{\mathbf{v}}_{\mathbf{y}}& \mathbf{=}& \frac{\mathbf{dy}}{\mathbf{dt}}\\ & \mathbf{=}& \frac{\mathbf{d}}{\mathbf{dt}}\mathbf{\left(}\mathbf{12}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{t}\mathbf{\right)}\end{array}$

vy = 24t - 2

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###### Problem Details

A particle's trajectory is described by x =(12t3−2t2)m and y =(12t2−2t)m, where t is in s.

Part A What is the particle's speed at t=0s ? v = m/s

What scientific concept do you need to know in order to solve this problem?

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