Question

A particle's trajectory is described by x =(12t3−2t2)m and y =(12t2−2t)m, where t is in s.

Answer 1

The particle's position is described by:

$x=(12t^3 -2t^2) m$

$y=(12t^2-2t)m$

1) particle's speed at t=0 s

To find the particle's velocity, we must differentiate the position in x,y:

$v_x = x' = (36 t^2 -4t) m/s$

$v_y = y' =(24 t-2) m/s$

These are the velocities along the two directions x and y. Substituting t=0 s, we find:

$v_x =0$

$v_y=-2 m/s$

So, the speed of the particle at t=0 is given by

$v=\sqrt{v_x^2+v_y^2}=\sqrt{0^2+(-2)^2}=2 m/s$

2)  particle's speed at t=5.0

we can use the equations we derived in the previous part and substitute t=5 s:

$v_x = 36 (5)^2 - 4 (5)= 880 m/s$

$v_y =24 (5) -2 =118 m/s$

So, the speed of the particle at t=5 is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(880)^2+(118)^2}=888.9 m/s$

3) particle's direction of motion, measured as an angle from the x-axis, at t=0 s

In this case, we see that the particle at t=0 s is moving toward the negative x-axis (it has no component of the velocity on the x-axis, while its vertical velocity is negative), so its direction is given by the angle

$\theta=270^{\circ}$

which corresponds to the negative vertical direction.

4) particle's direction of motion, measured as an angle from the x-axis, at t=5 s

The particle's direction of motion is given by

$\theta = arctan (\frac{v_y}{v_x})$

calculated at t=5 s, we have

$\theta = arctan(\frac{118}{880})= arctan(0.134)=7.64^{\circ}$

Answer 2

• 2 m/s
• 890 m/s
• 270°
• 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 $\boxed{ \ \vec{v} = (36t^2 - 4t) \hat{i} + (24t - 2) \hat{j} \ m/s \ }$

- - - - - - -

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.

- - - - - - -

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