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AIEEE 2005 Physics Solved Paper
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© examsnet.com
Question : 4
Total: 74
The relation between time
t
and distance
x
is
t
=
ax
2
+
bx
where
a
and
b
are constants. The acceleration is
2
bv
3
−
2
abv
3
2
av
3
−
2
av
3
Validate
Solution:
Given
t
=
a
x
2
+
b
x
Differentiate with respect to time
(
t
)
d
d
t
(
t
)
=
a
d
d
t
(
x
2
)
+
b
d
x
d
t
=
a
.2
x
d
x
d
t
+
b
⋅
d
x
d
t
1
=
2
a
x
v
+
b
v
=
v
(
2
a
x
+
b
)
[as
d
x
d
t
=
v
]
⇒
2
a
x
+
b
=
1
v
Again differentiating,
2
a
d
x
d
t
+
0
=
−
1
v
2
d
v
d
t
⇒
d
v
d
t
=
f
=
−
2
a
v
3
(as
d
v
d
t
=
f
=
Acceleration)
© examsnet.com
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