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A car starts from rest. It has to cover a distance of 500 m. the coefficient of friction between the road and the tyre is 0.5. The minimum time in which the car can cover this distance is (g=10m/ s2 ).
dear kaushik ram, the car moves due to the frictional force provided due to the motion of tyres in opposite direction, the frictional force = μmg therefore acceleration provided due to friction is 0.5*10 and using second law of motion, s=ut + 1/2*a*t2 t=2001/2sec. We are all IITians and here to help you in your IIT JEE preparation. Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar : Click here to download the toolbar We are all IITians and here to help you in your IIT JEE preparation. Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar : Click here to download the toolbar AMAN BANSAL
dear kaushik ram,
the car moves due to the frictional force provided due to the motion of tyres in opposite direction,
the frictional force = μmg
therefore acceleration provided due to friction is 0.5*10
and using second law of motion,
s=ut + 1/2*a*t2
t=2001/2sec.
We are all IITians and here to help you in your IIT JEE preparation.
Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar : Click here to download the toolbar
AMAN BANSAL
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