Given two vectors a bar is equal to x I cap minus 4 j cap and b bar is equal to 6 I cal plus 2 j cap for value of x match the following
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Answer:
=(t
2
−4t+6)
i
^
+(t
2
)
j
^
velocity =
dt
dr
=(2t−4)
i
^
+(2t)
j
^
acceleration =
dt
dv
=(2)
i
^
+(2)
j
^
a
.
v
=((2t−4)
i
^
+(2t)
j
^
).((2)
i
^
+(2)
j
^
)
when they are perpendicular then a.v = 0
((2t−4)
i
^
+(2t)
j
^
).((2)
i
^
+(2)
j
^
)=0
(2t−4)×2+4t=0
4t−8+4t=0
t=1s
solution
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