Question

A particle moves on the XY plane such that its position is given by
t ^{3}-3t and y = t ^{2}-2t for t>= 0,

where x and y are in meters and the time t is in seconds.

a) Find the magnitude and the direction (0 degree le theta < 360 degree) of the resultant velocity when t = 0.5 .

b) Does the particle ever come to a stop? If so, when and where?

Answer 1

\[ velocity = \sqrt{Vx^2 + Vy^2}\]

Answer 2

\[lvl=\sqrt{(3(0.5^{2}-3)^{2}+(2(0.5)-2)^{2}}\]

Answer 3

vx=3t^2-3
vy=2t-2

Answer 4

\[\tan \theta = \frac{ Vy}{ Vx }\]

Answer 5

\[lvl=2.4622 \tan−1(-1/-2.25)=23.9624degree\]

Answer 6

for part B how??

Answer 7

the particle will come to a stop when magnitude of velocity =0
use this fact for (B)

Answer 8

any further explanation for part B?

Answer 9

the particle comes to a full stop when both horizontal and velocities are equal to zero.

set \(v_x=0\) and \(v_y=0\) and solve for \(t\) in each equation. if there is a common value for \(t\), then at that instant, the particle is at rest.

Answer 10

\[3t ^{2}-3=0 and 2t-2=0?\]

Answer 11

thats right

Answer 12

\[3t ^{2}−3=0\]
\[t ^{3}=3t\]
\[\frac{ t ^{3} }{ t }=3\]<---- right?

Answer 13

nope.
\(3t^2-3=0\)

\(3t^2=3\)

\(t^2=1\)

\(t=1\)



\(2t-2=0\)

\(t=1\)

therefore, at \(t=1\) the particle is at rest.

Answer 14

that all it asking?

Answer 15

yes. find the time \(t\) such that the particle is at rest.

also, find its position when it is at rest.
put \(t=1\) in the equation for \(x\) and \(y\) and you'll get the coordinate of the point where it is at rest.

Answer 16

x=t^3-3t
=1^3-3(1)
=1-3
=-2

y=t^2-2t
=1^2-2(1)
=1-2
=-1

Answer 17

my answer are - ve not sure if it correct

Answer 18

negative values for \(x\) or \(y\) indicate that their position is to the left or below the origin.

Answer 19

x=-2m and y=-1m ?