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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.

Respuesta :

[tex]\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+\frac{at^2}{2} \\ x_f=\frac{at^2}{2} \\ x_f=\frac{(-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}[/tex]

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