A point moves along a straight line in such a way
That after t seconds its distance from origin is s=2t^2+3t meters
i)find average veleity of the points between t=3, t=6seconds
ii) Find instantaneous velocities at t=3, t=6 seconds
Answers
Answered by
0
Answer:
Answer:
∫
6
−
3
v
(
t
)
d
t
=
103.5
Step-by-step explanation:
Explanation:
The area under a velocity curve is equivalent to the distance covered.
∫
6
−
3
v
(
t
)
d
t
=
∫
6
−
3
−
t
2
+
3
t
−
2
X
d
t
=
−
1
3
t
3
+
3
2
t
2
−
2
t
∣
∣
∣
6
(
−
3
)
=
(
−
1
3
(
6
3
)
+
3
2
(
6
2
)
−
2
(
6
)
)
−
(
−
1
3
(
−
3
)
3
+
3
2
(
−
3
)
2
−
2
(
−
3
)
)
=
114
−
10.5
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