Physics, 10.12.2019 02:31, fy1lqjh

Aparticle's trajectory is described by x =(12t3−2t2)m and y =(12t2−2t)m, where t is in s.
what is the particle's speed at t=0s ? what is the particle's speed at t=5.0s ? express your answer using two significant figures. what is the particle's direction of motion, measured as an angle from the x-axis, at t=0 s ?
express your answer using two significant figures. what is the particle's direction of motion, measured as an angle from the x-axis, at t=5.0s ?
express your answer using two significant figures.

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answered: nmooesy2229

2 m/s890 m/s270°7.6°Further explanation

Given:

A particle's trajectory is described by

$\boxed{ \ x = (12t^3 - 2t^2) \ m \ }, and$ $\boxed{ \ y = (12t^2 - 2t) \ m \ }$

We can see that the particle moves in the direction in both the x-axis and the y-axis.

The instantaneous velocity component is obtained from the first derivative of the position function with respect to time.

$\boxed{ \ v_x = \frac{dx}{dt} \ } \ and \ \boxed{ \ v_y = \frac{dy}{dt} \ }$

Let's arrange the velocity function for horizontal and vertical components.

$\boxed{ \ v_x = 36t^2 - 4t \ m/s \ }$ $\boxed{ \ v_y = 24t - 2 \ m/s \ }$

If asked the vector of the velocity, then $\boxed{ \ \vec{v} = (36t^2 - 4t) \hat{i} + (24t - 2) \hat{j} \ m/s \ }$

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Determine the particle's speed at t = 0 s.

$t = 0 \rightarrow \boxed{ \ v_x = 36(0)^2 - 4(0) \ } \rightarrow \boxed{ \ v_x = 0 \ m/s \ }$

$t = 0 \rightarrow \boxed{ \ v_y = 24(0) - 2 \ } \rightarrow \boxed{ \ v_y = -2 \ m/s \ }$

In vector form, we can write it as $\boxed{ \ \vec{v} = - 2 \hat{j} \ m/s \ }$

From the two velocity components above, we calculate the result.

$\boxed{ \ v = \sqrt{v_x^2 + v^2_y} \ } \rightarrow \boxed{ \ v = \sqrt{0^2 + (-2)^2} \ }$

Hence, the speed of the particle at t = 0 is given by $\boxed{ \ v = 2 \ m/s \ }$

Determine the particle's speed at t = 5 s.

$t = 5 \rightarrow \boxed{ \ v_x = 36(5)^2 - 4(5) \ } \rightarrow \boxed{ \ v_x = 880 \ m/s \ }$

$t = 5 \rightarrow \boxed{ \ v_y = 24(5) - 2 \ } \rightarrow \boxed{ \ v_y = 118 \ m/s \ }$

In vector form, we can write it as $\boxed{ \ \vec{v} = 880 \hat{i} + 118 \hat{j} \ m/s \ }$

From the two velocity components above, we calculate the result.

$\boxed{ \ v = \sqrt{v_x^2 + v^2_y} \ } \rightarrow \boxed{ \ v = \sqrt{880^2 + 118^2} = 887.8761175 \ }$

Hence, the speed of the particle at t = 5 is given by $\boxed{ \ v = 890 \ m/s \ }$ in two significant figures.

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Now we determine the particle's direction of motion, measured as an angle from the x-axis, at t = 0 s and t = 5 s.

The direction of motion of a particle is given by an α angle which can be calculated by the trigonometric formula.

$\boxed{ \ tan \ \alpha = \frac{v_y}{v_x} \ } \rightarrow \boxed{ \ \alpha = arctan\frac{v_y}{v_x} \ }$

At t = 0 s,

$\boxed{ \ v_x = 0 \ } \ m/s$ $\boxed{ \ v_y = -2 \ } \ m/s$

$\boxed{ \ \alpha = arctan\frac{-2}{0} \ } \rightarrow \boxed{ \ \alpha = arctan( - \infty ) \ }$

Actually, from the two components of velocity, we can see the particle's direction of motion.

$\boxed{ \ v_x = 0 \ } \ m/s$ → it has no component. $\boxed{ \ v_y = -2 \ } \ m/s$ → move along in the negative y-axis direction.

Therefore the particle's direction of motion at t = 0 s is given by the angle $\boxed{ \ \alpha = 270^0 \ }$

At t = 5 s,

$\boxed{ \ v_x = 880 \ } \ m/s$ $\boxed{ \ v_y = 118 \ } \ m/s$

$\boxed{ \ \alpha = arctan\frac{118}{880} \ } \rightarrow \boxed{ \ \alpha = 7.6^0 \ }$ in two significant figures.

Therefore the particle's direction of motion at t = 5 s is given by the angle $\boxed{ \ \alpha = 7.6^0 \ }$

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Keywords: a particle's trajectory, speed, express your answer using two significant figures, the direction of motion, derivative, the instantaneous velocity component, horizontal, vertical, vector, the angle