A bird moves from point (1 m,-2 m, 3 m) to (4 m, 2 m, 3 m). If the speed of the bird is 10 m/s, then the velocity vector of the bird in m/s is:
(a) 5(i – 2j+3k)
(©) 0.61 +0.8ị
(b) 5(4 +2j+3k)
(d) 6î+8)
please explain with whole theory...I will mark you brainliest.
Answers
Answered by
1
Explanation:
Initial position vector of the bird
→
r
i
=
ˆ
i
−
2
ˆ
j
+
3
ˆ
k
Final position vector of the bird
→
r
f
=
4
ˆ
i
+
2
ˆ
j
+
3
ˆ
k
So displacement vector
→
d
=
→
r
f
−
→
r
i
=
(
4
ˆ
i
+
2
ˆ
j
+
3
ˆ
k
)
−
(
ˆ
i
−
2
ˆ
j
+
3
ˆ
k
)
=
3
ˆ
i
+
4
ˆ
j
Unit displacement vector
=
→
d
∣
∣
∣
→
d
∣
∣
∣
=
3
ˆ
i
+
4
ˆ
j
∣
∣
3
ˆ
i
+
4
ˆ
j
∣
∣
=
3
ˆ
i
+
4
ˆ
j
√
3
2
+
4
2
=
3
ˆ
i
+
4
ˆ
j
5
The direction of velocity is same as that of displacement.
And the magnitude of velocity is
10
m
/
s
Hence the velocity vector
→
v
=
10
×
3
ˆ
i
+
4
ˆ
j
5
m/s
⇒
→
v
=
6
ˆ
i
+
8
ˆ
j
m/s
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