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A particle passes through the origin with a velocity of (6.2m/s)y. If the particles acceleration is (-4.4m/s2)x, (a) What are its x and y positions after 5.0s? (b) What are Vx and Vy at this time? (c) Does the speed of this particle increase with time, decrease with time or increase and decrease? Explain.

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\begin{gathered} v_(ox)=0\text{ m/s} \\ v_(oy)=6.2\text{ m/s} \\ a_x=-4.4m/s^2 \\ t=5.0\text{ s} \\ a) \\ x-\text{position} \\ x_o=0\text{ m} \\ x_f=x_o+v_(ox)t+(at^2)/(2) \\ x_f=(at^2)/(2) \\ x_f=((-4.4m/s^2)(5.0s)^2)/(2) \\ x_f=-55\text{ m} \\ \text{The x-position is -55m} \\ y-\text{position} \\ y_f=v_(oy)t \\ y_f=(6.2\text{ m/s})(5.0\text{ s}) \\ y_f=31\text{ m} \\ \text{The y-position is 31m} \\ b) \\ \text{For v}_(fx) \\ \text{v}_(fx)=v_(ox)+a_xt \\ \text{v}_(fx)=a_xt \\ \text{v}_(fx)=(-4.4m/s^2)(5.0s) \\ \text{v}_(fx)=-22\text{ m/s} \\ \text{The v}_(fx)\text{ at that time is -22 m/s} \\ \text{For v}_(fy) \\ \text{ v}_(fy)=v_(oy)=6.2\text{ m/s} \\ \text{The v}_(fy)\text{ at that time is 6.2 m/s because it has constant velocity} \\ c) \\ The\text{ x-spe}ed\text{ increases with time at x-negative. Hence the resultant spe}ed\text{ increases too} \\ \text{with time} \end{gathered}

A particle passes through the origin with a velocity of (6.2m/s)y. If the particles-example-1
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