HC Verma Volume-1 Ch.-3 Kinematics Q.31. Can someone give me the solution?
HC Verma Volume-1 Ch.-3 Kinematics Q.31. Can someone give me the solution?
Shrey , 10 Years ago
Grade 11
Discuss with Askiitians Tutors> HC Verma Volume-1 Ch.-3 Kinematics Q.31. ...
3 Answerschetan mannan
Last Activity: 10 Years ago
31. For elevator and coin u = 0
As the elevator descends downward with acceleration a (say)
The coin has to move more distance than 1.8 m to strike the floor. Time taken t = 1 sec.
1 ut = 0 + 1/2 g(1)2 = 1/2 g
Sc = 2
1 ut = u + 1/2 a(1)2 = 1/2 a
Se = 2 at
Total distance covered by coin is given by = 1.8 + 1/2 a = 1/2 g
1.8 +a/2 = 9.8/2 = 4.9
a = 6.2 m/s2 = 6.2 × 3.28 = 20.34 ft/s2
ta
2
2
a 6ft=1.8m
31. For elevator and coin u = 0
As the elevator descends downward with acceleration a (say)
The coin has to move more distance than 1.8 m to strike the floor. Time taken t = 1 sec.
1 ut = 0 + 1/2 g(1)2 = 1/2 g
Sc = 2
1 ut = u + 1/2 a(1)2 = 1/2 a
Se = 2 at
Total distance covered by coin is given by = 1.8 + 1/2 a = 1/2 g
1.8 +a/2 = 9.8/2 = 4.9
a = 6.2 m/s2 = 6.2 × 3.28 = 20.34 ft/s2
ta
2
2
a 6ft=1.8m
31. For elevator and coin u = 0
As the elevator descends downward with acceleration a (say)
The coin has to move more distance than 1.8 m to strike the floor. Time taken t = 1 sec.
1 ut = 0 + 1/2 g(1)2 = 1/2 g
Sc = 2
1 ut = u + 1/2 a(1)2 = 1/2 a
Se = 2 at
Total distance covered by coin is given by = 1.8 + 1/2 a = 1/2 g
1.8 +a/2 = 9.8/2 = 4.9
a = 6.2 m/s2 = 6.2 × 3.28 = 20.34 ft/s2
ta
2
2
a 6ft=1.8m
31. For elevator and coin u = 0
As the elevator descends downward with acceleration a (say)
The coin has to move more distance than 1.8 m to strike the floor. Time taken t = 1 sec.
1 ut = 0 + 1/2 g(1)2 = 1/2 g
Sc = 2
1 ut = u + 1/2 a(1)2 = 1/2 a
Se = 2 at
Total distance covered by coin is given by = 1.8 + 1/2 a = 1/2 g
1.8 +a/2 = 9.8/2 = 4.9
a = 6.2 m/s2 = 6.2 × 3.28 = 20.34 ft/s2
ta
2
2
a 6ft=1.8m
31. For elevator and coin u = 0
As the elevator descends downward with acceleration a (say)
The coin has to move more distance than 1.8 m to strike the floor. Time taken t = 1 sec.
1 ut = 0 + 1/2 g(1)2 = 1/2 g
Sc = 2
1 ut = u + 1/2 a(1)2 = 1/2 a
Se = 2 at
Total distance covered by coin is given by = 1.8 + 1/2 a = 1/2 g
1.8 +a/2 = 9.8/2 = 4.9
a = 6.2 m/s2 = 6.2 × 3.28 = 20.34 ft/s2
ta
2
2
a 6ft=1.8m
31. For elevator and coin u = 0
As the elevator descends downward with acceleration a (say)
The coin has to move more distance than 1.8 m to strike the floor. Time taken t = 1 sec.
1 ut = 0 + 1/2 g(1)2 = 1/2 g
Sc = 2
1 ut = u + 1/2 a(1)2 = 1/2 a
Se = 2 at
Total distance covered by coin is given by = 1.8 + 1/2 a = 1/2 g
1.8 +a/2 = 9.8/2 = 4.9
a = 6.2 m/s2 = 6.2 × 3.28 = 20.34 ft/s2
ta
2
2
a 6ft=1.8m
31. For elevator and coin u = 0
As the elevator descends downward with acceleration a (say)
The coin has to move more distance than 1.8 m to strike the floor. Time taken t = 1 sec.
1 ut = 0 + 1/2 g(1)2 = 1/2 g
Sc = 2
1 ut = u + 1/2 a(1)2 = 1/2 a
Se = 2 at
Total distance covered by coin is given by = 1.8 + 1/2 a = 1/2 g
1.8 +a/2 = 9.8/2 = 4.9
a = 6.2 m/s2 = 6.2 × 3.28 = 20.34 ft/s2
ta
2
2
a 6ft=1.8m
31. For elevator and coin u = 0
As the elevator descends downward with acceleration a (say)
The coin has to move more distance than 1.8 m to strike the floor. Time taken t = 1 sec.
1 ut = 0 + 1/2 g(1)2 = 1/2 g
Sc = 2
1 ut = u + 1/2 a(1)2 = 1/2 a
Se = 2 at
Total distance covered by coin is given by = 1.8 + 1/2 a = 1/2 g
1.8 +a/2 = 9.8/2 = 4.9
a = 6.2 m/s2 = 6.2 × 3.28 = 20.34 ft/s2
ta
2
2
a 6ft=1.8m
31. For elevator and coin u = 0
As the elevator descends downward with acceleration a (say)
The coin has to move more distance than 1.8 m to strike the floor. Time taken t = 1 sec.
1 ut = 0 + 1/2 g(1)2 = 1/2 g
Sc = 2
1 ut = u + 1/2 a(1)2 = 1/2 a
Se = 2 at
Total distance covered by coin is given by = 1.8 + 1/2 a = 1/2 g
1.8 +a/2 = 9.8/2 = 4.9
a = 6.2 m/s2 = 6.2 × 3.28 = 20.34 ft/s2
ta
2
2
a 6ft=1.8m
31. For elevator and coin u = 0
As the elevator descends downward with acceleration a (say)
The coin has to move more distance than 1.8 m to strike the floor. Time taken t = 1 sec.
1 ut = 0 + 1/2 g(1)2 = 1/2 g
Sc = 2
1 ut = u + 1/2 a(1)2 = 1/2 a
Se = 2 at
Total distance covered by coin is given by = 1.8 + 1/2 a = 1/2 g
1.8 +a/2 = 9.8/2 = 4.9
a = 6.2 m/s2 = 6.2 × 3.28 = 20.34 ft/s2
ta
2
2
a 6ft=1.8m
31. For elevator and coin u = 0
As the elevator descends downward with acceleration a (say)
The coin has to move more distance than 1.8 m to strike the floor. Time taken t = 1 sec.
1 ut = 0 + 1/2 g(1)2 = 1/2 g
Sc = 2
1 ut = u + 1/2 a(1)2 = 1/2 a
Se = 2 at
Total distance covered by coin is given by = 1.8 + 1/2 a = 1/2 g
1.8 +a/2 = 9.8/2 = 4.9
a = 6.2 m/s2 = 6.2 × 3.28 = 20.34 ft/s2
ta
2
2
a 6ft=1.8m
Gayatri Jayesh Bondriya
Last Activity: 10 Years ago
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Soneel Verma
Last Activity: 10 Years ago
You may consult IE Irodov book for solutions and for advanced practice and for the sake of looking for reputed institutions, you may join any of these – BHU, Varanasi,...........................TKM College of Engg, Kollam,..................Gaya College of Engg, GAya,....................Lovely Professional Uni, Phagwara
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