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JEE Main 25 July 2021 Shift 2 Solved Paper
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© examsnet.com
Question : 1
Total: 90
The relation between time
t
and distance
x
for a moving body is given as
t
=
m
x
2
+
n
x
, where
m
and
n
are constants. The retardation of the motion is: (When v stands for velocity)
[25 Jul 2021 Shift 2]
2
m
v
3
2
m
n
v
3
2
n
v
3
2
n
2
v
3
Validate
Solution:
t
=
m
x
2
+
n
x
1
v
=
d
t
d
x
=
2
m
x
+
n
v
=
1
2
m
x
+
n
d
v
d
t
=
−
2
m
(
2
m
x
+
n
)
2
(
d
x
d
t
)
a
=
−
(
2
m
)
v
3
© examsnet.com
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