a car tavelling at 9ms-1 accelerate and attain a speed of 27ms-1 in5s. calculate the acceleration and the distance covered in 5s
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U=9 m/s
V=27m/s
T=5s
By first law of motion
V=u+at
27=9+a*5
5a=27-9
A=18/5
A=3.6answer
Hope it may help you
V=27m/s
T=5s
By first law of motion
V=u+at
27=9+a*5
5a=27-9
A=18/5
A=3.6answer
Hope it may help you
abhishekchaudhary74:
like please
and gave comments also
but iska answer 3.6 aur 90 hai
sorry I forgot that
yaar thanks please
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