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JEE Main Physics Class 11 Motion in a Straight Line and Plane Part 3 Questions
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© examsnet.com
Question : 40
Total: 45
From the
v
−
t
graph shown, the ratio of distance to displacement in
25
s
of motion is:
[11-Apr-2023 shift 1]
3
5
1
2
5
3
1
Validate
Solution:
Displacement
=
Area of graph with sign
Displacement
=
(
1
2
×
10
×
5
)
+
(
10
×
5
)
+
(
1
2
×
5
×
30
)
+
(
1
2
×
5
×
20
)
−
1
2
(
5
)
(
20
)
=
25
+
50
+
75
+
50
−
50
=
150
m
Distance
→
Area of graph with positive value
Distance
=
25
+
50
+
75
+
50
=
250
Distance
Displacement
=
250
150
=
5
3
option
→
(
3
)
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