A particle moves in xy plane under acceleration a=3i +4jm/s².if the initial velocity is u=6i+8jm/s.find the velocity and the position vector of particle afterc2s
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Answer:
Initial velocity of particle
u
= 3
i
^
+4
j
^
Acceleration
a
= 4
i
^
−3
j
^
Final velocity after 1 second
v
=
u
+
a
t
where t=1 s
⟹
v
= (3
i
^
+4
j
^
)+ (4
i
^
−3
j
^
)×1= 7
i
^
+
j
^
Speed after 1 second ∣
v
∣=
7
2
+1
=
50
m/s
Explanation:
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