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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?

Grade:11

3 Answers

chetan mannan
13 Points
8 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
521 Points
8 years ago
      
 
                     it will help you      ….….….…..PLEASE   ..... if    u   like  my  answer  and  my  answer   is  use  full    for  you  so please   just click on approve  button  ...and  if  you  have  any  other  question   or  dought     keep asking
Soneel Verma
823 Points
8 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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