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The position of an electron is given by r = 7.26t - 6.03t2 + 6.41, with t in seconds and r in meters. At t = 3.03 s, what are (a) the x-component, (b) the y-component, (c) the magnitude, and (d) the angle relative to the positive direction of the x axis, of the electron's velocity ?

User Jmettraux
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1 Answer

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Final answer:

The x-component of the electron's velocity is -28.7022 m/s, the y-component is -36.5418 m/s, the magnitude is 46.0119 m/s, and the angle relative to the positive direction of the x-axis is approximately 51.4 degrees.

Step-by-step explanation:

To find the x-component of the velocity, we can differentiate the position function r with respect to time t. Taking the derivative of r with respect to t gives us the velocity function v = 7.26 - 12.06t. Plugging in t = 3.03 s, we get v = 7.26 - 12.06(3.03) = -28.7022 m/s. Therefore, the x-component of the velocity is -28.7022 m/s.

Similarly, to find the y-component of the velocity, we differentiate y = -6.03t^2 with respect to t, which gives us vy = -12.06t. Plugging in t = 3.03 s, we get vy = -12.06(3.03) = -36.5418 m/s. Therefore, the y-component of the velocity is -36.5418 m/s.

To find the magnitude of the velocity, we can use the formula magnitude = sqrt(vx^2 + vy^2), where vx and vy are the x and y-components of the velocity respectively. Plugging in the values we found earlier, we get magnitude = sqrt((-28.7022)^2 + (-36.5418)^2) = 46.0119 m/s. Therefore, the magnitude of the velocity is 46.0119 m/s.

Lastly, to find the angle relative to the positive direction of the x-axis, we can use the formula angle = arctan(vy/vx). Plugging in the values we found earlier, we get angle = arctan((-36.5418)/(-28.7022)) = arctan(1.2743) = 0.896 radians (approximately 51.4 degrees). Therefore, the angle relative to the positive direction of the x-axis is approximately 51.4 degrees.

User Pototo
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