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JEE Main 9 Jan 2020 Shift 1 Solved Paper
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© examsnet.com
Question : 23
Total: 75
The distance x covered by a particle in one dimensional motion varies with time t as
x
2
=
a
t
2
+
2
b
t
+
c
. If the acceleration of the particle depends on
x
as
x
−
n
, where n is an integer, the value of n is
Your Answer:
Validate
Solution:
x
=
√
a
t
2
+
2
b
t
+
c
Differentiating w.r.t. time
d
x
d
t
=
v
=
1
2
√
a
t
2
+
2
b
t
+
c
×
(
2
a
t
+
2
b
)
⇒
v
=
a
t
+
b
x
⇒
v
x
=
a
t
+
b
Differentiating w.r.t. x
⇒
d
v
d
x
×
x
+
v
=
a
×
d
t
d
x
Multiply both side by v
⇒
(
v
d
v
d
x
)
x
+
v
2
=
a
⇒
a
′
x
=
a
−
v
2
[Here a' is acceleration]
⇒
a
′
x
=
a
−
(
a
t
+
b
x
)
2
⇒
a
′
x
=
a
x
2
−
(
a
t
+
b
)
2
x
2
⇒
a
′
x
=
a
(
a
t
2
+
2
b
t
+
c
)
−
(
a
t
+
b
)
2
x
2
⇒
a
′
=
a
c
−
b
2
x
3
∴
a
′
∝
1
x
3
∴
n
=
3
© examsnet.com
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