A man accelerates on a straight road from rest to a speed of 100km/hr in 1min. assuming uniform acceleration of man through out , what is the distance covered in this time
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ANSWER :
Given :
▪ Initial speed = zero (i.e. rest)
▪ Final speed = 100kmph
▪ Time interval = 1min
To Find :
▪ Distance covered in the given interval of time.
Concept :
▪ Since, acceleration is constant throughout the whole journey, we can apply equations of kinematics directly.
▪ First, we have to find out acceleration after that we can calculate distance with the help of second equation of kinematics.
▪ Acceleration is defined as the ratio of change in velocity to the time interval.
▪ Acceleration is a vector quantity.
▪ It can be positive, negative and zero.
Mathematically,
☞ a = (v - u)/t
Conversion :
✒ 1kmph = 5/18mps
✒ 100kmph = 100×5/18 = 27.78mps
✒ 1min = 60s
Calculation :
→ v^2 - u^2 = 2as
→ v^2 - u^2 = 2(v - u/t)s
Initial velocity = 0
→ v^2 = 2vs/t
→ v = 2s/t
→ s = vt/2
→ s = 27.78×60/2
→ s = 1666.8/2
→ s = 833.4m
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