describe the motion of body which is accelerating at a constant rate of 10ms-2 if the bady starts from rest how much distance will it cover in 2secs
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We have a formula
s = ut + 1/2at^2
u is the initial velocity which is equal to 0
a is the acceleration which is 10ms^-2
t is the time which is 2second
s is the distance
s = 0×2 + 1/2×10×2^2
s= 0+ 5×4
s = 20metre
therefore the distance covered will be 20 metre
s = ut + 1/2at^2
u is the initial velocity which is equal to 0
a is the acceleration which is 10ms^-2
t is the time which is 2second
s is the distance
s = 0×2 + 1/2×10×2^2
s= 0+ 5×4
s = 20metre
therefore the distance covered will be 20 metre
nini9:
no its ok
i dont need it now i got it
but whear are you from
Tamil nadu
ok thanks
you are very nice
are you on insta
Thnks
No
ok then bye
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