Question

0.334 kg particle moves in an xy plane according to x(t)=−10.0+2.00t−3.00t3 and y(t)=22.0+5.00t−10.0t2, with x and y in neters and t in seconds. At t=1.91

Answer 1

To find the position of a particle in an xy plane at a particular time, substitute the given time into the x(t) and y(t) functions. These calculations will give the x-coordinate and y-coordinate, respectively, indicating the particle's location at that time.

Step-by-step explanation:

The numerical values in your question suggest a particle's position in the xy plane over time, according to the equations x(t)=−10.0+2.00t−3.00t^3 and y(t)=22.0+5.00t−10.0t^2, with t=1.91. Now, to find the position of the particle at a particular time, we substitute the value of time (t) in the given equations. Here, time (t) is 1.91.

So, x(1.91) = -10.0 + 2.00*1.91 - 3.00*(1.91)^3 and y(1.91) = 22.0 + 5.00*1.91 - 10.0*(1.91)^2. After calculating these expressions, you'll find the x-coordinate and y-coordinate where the particle resides at time t=1.91.

Remember, in these equations, x(t) and y(t) denote the particle's position along the x-axis and y-axis respectively over time. Thus, they're called the particle's position functions.

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