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Starting from rest a scooter acquires a velocity of 36 km per hour in 10 seconds and then brakes are applied it takes 25 seconds to stop find out the acceleration and distance travelled during this motion

Posted by Mayank Som 1 month, 3 weeks ago

CBSE > Class 09 > Science

2 answers

Mohd Rihan 1 month, 3 weeks ago

Find out the story be hind the discovery of nuclear and protoplasm and write it in brief

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Kushagra Goyal 1 month, 3 weeks ago

Case I Initial velocity (u)=0m/s Final velocity (v)= 36km/h=36×5/18= 10m/s [1km/h=5/18 m/s] Time= 20 sec Acceleration=? By using first equation of motion [V=u+at] 36 =0 + a×20 36 - 0 = 20a 36/20 =a 9/5 = a 1.8m/s² Case II Initial velocity (u)=36km/h= 36×5/18= 10m/s [1km/h=5/18 m/s] Final velocity (v)= 0m/s Time= 25 sec Acceleration=? By using first equation of motion [V=u+at] 0= 36 + a×25 0-36=25a -36/25=a 1.44m/s² Case I Initial velocity (u)=0m/s Final velocity (v)= 36km/h=36×5/18= 10m/s [1km/h=5/18 m/s] Time= 20 sec Acceleration=1.8m/s² Distance=? By using second equation of motion [s=ut+½at²] 0×20+½×1.8×20² =360m Case II Initial velocity (u)=36km/h= 36×5/18= 10m/s [1km/h=5/18 m/s] Final velocity (v)= 0m/s Time= 25 sec Acceleration=1.44m/s² Distance=? By using second equation of motion [s=ut+½at²] 10×25+½1.44×25² =700m

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