Given:
A particle's trajectory is described by
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.
[tex]\boxed{ \ v_x = \frac{dx}{dt} \ } \ and \ \boxed{ \ v_y = \frac{dy}{dt} \ }[/tex]
Let's arrange the velocity function for horizontal and vertical components.
If asked the vector of the velocity, then [tex]\boxed{ \ \vec{v} = (36t^2 - 4t) \hat{i} + (24t - 2) \hat{j} \ m/s \ }[/tex]
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Determine the particle's speed at t = 0 s.
[tex] t = 0 \rightarrow \boxed{ \ v_x = 36(0)^2 - 4(0) \ } \rightarrow \boxed{ \ v_x = 0 \ m/s \ } [/tex]
[tex] t = 0 \rightarrow \boxed{ \ v_y = 24(0) - 2 \ } \rightarrow \boxed{ \ v_y = -2 \ m/s \ } [/tex]
In vector form, we can write it as [tex]\boxed{ \ \vec{v} = - 2 \hat{j} \ m/s \ }[/tex]
From the two velocity components above, we calculate the result.
[tex]\boxed{ \ v = \sqrt{v_x^2 + v^2_y} \ } \rightarrow \boxed{ \ v = \sqrt{0^2 + (-2)^2} \ }[/tex]
Hence, the speed of the particle at t = 0 is given by [tex]\boxed{ \ v = 2 \ m/s \ }[/tex]
Determine the particle's speed at t = 5 s.
[tex] t = 5 \rightarrow \boxed{ \ v_x = 36(5)^2 - 4(5) \ } \rightarrow \boxed{ \ v_x = 880 \ m/s \ } [/tex]
[tex] t = 5 \rightarrow \boxed{ \ v_y = 24(5) - 2 \ } \rightarrow \boxed{ \ v_y = 118 \ m/s \ } [/tex]
In vector form, we can write it as [tex]\boxed{ \ \vec{v} = 880 \hat{i} + 118 \hat{j} \ m/s \ }[/tex]
From the two velocity components above, we calculate the result.
[tex]\boxed{ \ v = \sqrt{v_x^2 + v^2_y} \ } \rightarrow \boxed{ \ v = \sqrt{880^2 + 118^2} = 887.8761175 \ }[/tex]
Hence, the speed of the particle at t = 5 is given by [tex]\boxed{ \ v = 890 \ m/s \ }[/tex] 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.
[tex]\boxed{ \ tan \ \alpha = \frac{v_y}{v_x} \ } \rightarrow \boxed{ \ \alpha = arctan\frac{v_y}{v_x} \ }[/tex]
At t = 0 s,
[tex]\boxed{ \ \alpha = arctan\frac{-2}{0} \ } \rightarrow \boxed{ \ \alpha = arctan( - \infty ) \ }[/tex]
Actually, from the two components of velocity, we can see the particle's direction of motion.
Therefore the particle's direction of motion at t = 0 s is given by the angle [tex]\boxed{ \ \alpha = 270^0 \ }[/tex]
At t = 5 s,
[tex]\boxed{ \ \alpha = arctan\frac{118}{880} \ } \rightarrow \boxed{ \ \alpha = 7.6^0 \ }[/tex] in two significant figures.
Therefore the particle's direction of motion at t = 5 s is given by the angle [tex]\boxed{ \ \alpha = 7.6^0 \ }[/tex]
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
