a point moves such that it displaces as a function of time given by x3=t3+1
Answers
Answered by
0
Answer:
Correct option is
B
x
3
2t
x
3
=t
3
+1
x=(t
3
+1)
1/3
The velocity is given by v=
dt
dx
=
dt
d
(t
3
+1)
1/3
=
3
1
(t
3
+1)
−2/3
.3t
2
v=t
2
.(t
3
+1)
−2/3
The acceleration is given by a=
dt
dv
=
dt
d
(t
2
(t
3
+1)
−2/3
)
=2t.(t
3
+1)
−2/3
+t
2
.
3
−2
(t
3
+1)
−5/3
.3t
2
=2t(x
3
)
−2/3
−2t
4
(x
3
)
−5/3
=2tx
−2
−2t
4
x
−5
=2t(x
−2
−t
3
.x
−5
)
=2t(
x
2
1
−
x
5
x
3
−1
)∵t
3
=x
3
−1
On solving
a=
x
3
2t
Answered by
0
Answer
Correct option is
B
x
3
2t
x
3
=t
3
+1
x=(t
3
+1)
1/3
The velocity is given by v=
dt
dx
=
dt
d
(t
3
+1)
1/3
=
3
1
(t
3
+1)
−2/3
.3t
2
v=t
2
.(t
3
+1)
−2/3
The acceleration is given by a=
dt
dv
=
dt
d
(t
2
(t
3
+1)
−2/3
)
=2t.(t
3
+1)
−2/3
+t
2
.
3
−2
(t
3
+1)
−5/3
.3t
2
=2t(x
3
)
−2/3
−2t
4
(x
3
)
−5/3
=2tx
−2
−2t
4
x
−5
=2t(x
−2
−t
3
.x
−5
)
=2t(
x
2
1
−
x
5
x
3
−1
)∵t
3
=x
3
−1
On solving
a=
x
3
2t
Please mark as Brainliest.
Correct option is
B
x
3
2t
x
3
=t
3
+1
x=(t
3
+1)
1/3
The velocity is given by v=
dt
dx
=
dt
d
(t
3
+1)
1/3
=
3
1
(t
3
+1)
−2/3
.3t
2
v=t
2
.(t
3
+1)
−2/3
The acceleration is given by a=
dt
dv
=
dt
d
(t
2
(t
3
+1)
−2/3
)
=2t.(t
3
+1)
−2/3
+t
2
.
3
−2
(t
3
+1)
−5/3
.3t
2
=2t(x
3
)
−2/3
−2t
4
(x
3
)
−5/3
=2tx
−2
−2t
4
x
−5
=2t(x
−2
−t
3
.x
−5
)
=2t(
x
2
1
−
x
5
x
3
−1
)∵t
3
=x
3
−1
On solving
a=
x
3
2t
Please mark as Brainliest.
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