IIT JEE , AIEEE Forum - Physics , Mechanics

tarun_bits

The change in velocity with distance is not constant- it decreases with increasing distance of fall,,,,,,explain how?

edison

Which distance and fall you are talking about.
Please expalin the conditions and situations you are referring to.

taruntanuj007

see while freefall only g isacting

now a = v dv/dx (derive it its easy)

since v is increasing per unti time with distance but remember that velocity increases increases linearly ie @ g per second but distance increases quadratically ie 1/2 g t². so obviously distance increases faster than velocity....as per equation given above dv/dx * v is a constant ie g in this case.And u can intuitively understand that when velocity has increased by g the distance has increased by much more than due to the square so dv/dx changes....emply these in equation of motion to get a cleearer picture!!!

INDIAN_ARMY19890

v=gt+u
differenciating w.r.t x
dv/dx=g(dt/dx)
now dt/dx is time taken to travel unit distance as u know velocity will goes on increasing during free fall hence dt/dx is a decreasing function
g is constant
hence dv/dx is a decreasing function

tarun_bits

tarun_bits

army....earlier u said that dt/dx is decreasing function then u said that dv/dx is decreasing

INDIAN_ARMY19890

OF COURSE YAAR AS
dv/dx=gdt/dx g is constant dt/dx is a decreasing function dv/dx is directly propotional to it that"s why dv/dx is also a decreasing function
i hope u got why dt/dx is decreasing if not then just ask

tarun_bits

thanks

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