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Motion in a Plane

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Question : 20
Total: 32
The position of a particle is given by r=3.0t
^
i
−2.0t2
^
j
+4.0
^
k
m where t is in seconds and the coefficients have the proper units for r to be in metres.
(a) Find the
→
v
 and
→
a
 of the particle?
(b) What is the magnitude and direction of velocity of the particle at t=2.0s?
Solution:  
(a) Velocity,
→
v
=
d
→
r
dt
 =
d
dt
(3.0t
^
i
−2.0t2
^
j
+4.0
^
k
) =[3.0
^
i
−4.0t
^
j
]ms−1
Acceleration,
→
a
=
d
→
v
dt
 =
d
dt
(3.0
^
i
−4.0t
^
j
)=0−4.0
^
j
ms−2
(b) At time t=2s,
→
v
=3.0
→
i
−4.0×2
→
j
 =3.0
→
i
−8.0
→
j

∴v=√(3.0)2+(−8)2=√73=8.54ms−1
If θ is the angle which makes with x-axis, then
tan‌θ=
vy
vx
=
−8
3
=−2.667
∴θ=tan−1(−2.667)≃70° with x -axis
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