Question

A body staring from rest. Travels with uniform acceleration if it 100m in5s, what is the value of acceleration ​

Answer 1

Answer :-

Initial velocity = u = 0

velocity = u = 0distance traveled = s = 100 m

velocity = u = 0distance traveled = s = 100 mTime = t = 5 sec

velocity = u = 0distance traveled = s = 100 mTime = t = 5 sec a = acceleration

velocity = u = 0distance traveled = s = 100 mTime = t = 5 sec a = accelerationEquation of motion :-

velocity = u = 0distance traveled = s = 100 mTime = t = 5 sec a = accelerationEquation of motion :-s = ut + 1/2at²

velocity = u = 0distance traveled = s = 100 mTime = t = 5 sec a = accelerationEquation of motion :-s = ut + 1/2at² = 0 + 1/2 a × 5 × 5

velocity = u = 0distance traveled = s = 100 mTime = t = 5 sec a = accelerationEquation of motion :-s = ut + 1/2at² = 0 + 1/2 a × 5 × 5100 = 1/2a × 25

velocity = u = 0distance traveled = s = 100 mTime = t = 5 sec a = accelerationEquation of motion :-s = ut + 1/2at² = 0 + 1/2 a × 5 × 5100 = 1/2a × 251/2a = 100 / 25

velocity = u = 0distance traveled = s = 100 mTime = t = 5 sec a = accelerationEquation of motion :-s = ut + 1/2at² = 0 + 1/2 a × 5 × 5100 = 1/2a × 251/2a = 100 / 251/2a = 4

velocity = u = 0distance traveled = s = 100 mTime = t = 5 sec a = accelerationEquation of motion :-s = ut + 1/2at² = 0 + 1/2 a × 5 × 5100 = 1/2a × 251/2a = 100 / 251/2a = 4a = 4 × 2

velocity = u = 0distance traveled = s = 100 mTime = t = 5 sec a = accelerationEquation of motion :-s = ut + 1/2at² = 0 + 1/2 a × 5 × 5100 = 1/2a × 251/2a = 100 / 251/2a = 4a = 4 × 2 Answer = 8 m/s²