please help fast see the question in the photo
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Answer:
Acceleration during O→A
a
OA
=
20−0
30−0
=
20
30
=1.5m/s
2
during AB
a
AB
=0
during BC
a
BC
=
t
C
−t
0
v
C
−v
B
=
40−30
80−30
=
10
50
=5m/s
2
during CD
a
CD
=
t
D
−t
C
v
D
−v
C
=
80−40
0−80
=
40
−80
=−2m/s
2
maximum acceleration =5m/s
2
Explanation:
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