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

Asked by maham237 @ 22/12/2020 in Physics viewed by 412 People

A particle'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.

Answered by maham237 @ 22/12/2020

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

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

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