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Class 11 NEET Physics Motion in a Straight Line Questions
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© examsnet.com
Question : 47
Total: 58
A car accelerates from rest at a constant rate
α
for some time after which it decelerates at a constant rate
β
and comes to rest. If total time elapsed is
t
,
then maximum velocity acquired by car will be
( 1994 )
(
α
2
−
β
2
)
t
α
β
(
α
2
+
β
2
)
t
α
β
(
α
+
β
)
t
α
β
α
β
t
α
+
β
Validate
Solution:
Initial velocity
(
u
)
=
0
;
acceleration in the first phase
=
α
;
deceleration in the second phase
=
β
and total time
=
t
.
When car is accelerating then final velocity
(
v
)
=
u
+
α
t
=
0
+
α
t
1
or
t
1
=
v
α
and when car is decelerating,
then final velocity
0
=
v
−
β
t
or
t
2
=
v
β
Therefore total time
(
t
)
=
t
1
+
t
2
=
v
α
+
v
β
t
=
v
(
1
α
+
1
β
)
=
v
(
β
+
α
α
β
)
or
v
=
α
β
t
α
+
β
© examsnet.com
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