Abody moving in a straight line with uniform acceleration is found to travel 10 min 4 sec and 20 min the next 4 seconds. Find its acceleration and displacement in the 4th sec. interval.
Answers
Answered by
0
Correct option is
D
5
The distance covered in n
th
second,
S
n
=u+
2
a
(2n−1) . . . . . . . . . . .(1)
For n=3 and S
3
S
3
=u+
2
a
(6−1)=10
10=u+
2
5a
. . . . . . . . .(2)
For n=4 and S
4
=12m
S
4
=u+
2
a
(8−1)
12=u+
2
7a
. . . . . . . . .(3)
Subtract equation (3) by equation (2),
12−10=(u+
2
7a
)−(u+
2
5a
)
a=2m/s
2
From equation (2),
10=u+
2
5×2
10=u+5
u=5m/s
The correct option is D.
Answered by
0
T
1
=2π
g
1
T
2
=2π
g
9
=3×2π
g
1
=3T
1
therefore time period of second pendulum is thrice of first pendulum.
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