*18) The speed of a motor car is reduced from 54 km/hr to 36 km/hr in 15 second. Find the distance covered by the car during this time. How much farther will the car travel before coming to rest? Assume that the retardation is uniform throughout the motion.

*18) The speed of a motor car is reduced from 54 km/hr to 36 km/hr in

1.	*18) The speed of a motor car is reduced from 54 km/hr to 36 km/hr in 15 second. Find the distance covered by the car during this time. How much farther will the car travel before coming to rest? Assume that the retardation is uniform throughout the motion.
Answer» Here, Side of the given square field = 10mHere, Side of the given square field = 10mso, perimeter of a square = 4side = 10 m 4 = 40 mHere, Side of the given square field = 10mso, perimeter of a square = 4side = 10 m 4 = 40 mFarmer TAKES 40 s to move along the boundary.Here, Side of the given square field = 10mso, perimeter of a square = 4side = 10 m 4 = 40 mFarmer takes 40 s to move along the boundary.Displacement after 2 minutes 20 s = 2 * 60 s + 20 s = 140 secondsHere, Side of the given square field = 10mso, perimeter of a square = 4side = 10 m 4 = 40 mFarmer takes 40 s to move along the boundary.Displacement after 2 minutes 20 s = 2 * 60 s + 20 s = 140 secondssince in 40 s farmer moves 40 mHere, Side of the given square field = 10mso, perimeter of a square = 4side = 10 m 4 = 40 mFarmer takes 40 s to move along the boundary.Displacement after 2 minutes 20 s = 2 * 60 s + 20 s = 140 secondssince in 40 s farmer moves 40 mTherefore, in 1s the distance covered by farmer = 40 / 40 m = 1mHere, Side of the given square field = 10mso, perimeter of a square = 4side = 10 m 4 = 40 mFarmer takes 40 s to move along the boundary.Displacement after 2 minutes 20 s = 2 * 60 s + 20 s = 140 secondssince in 40 s farmer moves 40 mTherefore, in 1s the distance covered by farmer = 40 / 40 m = 1mTherefore, in 140s distance covered by farmer = 1 � 140 m = 140 m. Now, number of rotation to COVER 140 along the boundary= Total Distance / PerimeterHere, Side of the given square field = 10mso, perimeter of a square = 4side = 10 m 4 = 40 mFarmer takes 40 s to move along the boundary.Displacement after 2 minutes 20 s = 2 * 60 s + 20 s = 140 secondssince in 40 s farmer moves 40 mTherefore, in 1s the distance covered by farmer = 40 / 40 m = 1mTherefore, in 140s distance covered by farmer = 1 � 140 m = 140 m. Now, number of rotation to cover 140 along the boundary= Total Distance / Perimeter= 140 m / 40 m = 3.5 roundHere, Side of the given square field = 10mso, perimeter of a square = 4side = 10 m 4 = 40 mFarmer takes 40 s to move along the boundary.Displacement after 2 minutes 20 s = 2 * 60 s + 20 s = 140 secondssince in 40 s farmer moves 40 mTherefore, in 1s the distance covered by farmer = 40 / 40 m = 1mTherefore, in 140s distance covered by farmer = 1 � 140 m = 140 m. Now, number of rotation to cover 140 along the boundary= Total Distance / Perimeter= 140 m / 40 m = 3.5 roundThus, after 3.5 round farmer will at point C of the field.Here, Side of the given square field = 10mso, perimeter of a square = 4side = 10 m 4 = 40 mFarmer takes 40 s to move along the boundary.Displacement after 2 minutes 20 s = 2 * 60 s + 20 s = 140 secondssince in 40 s farmer moves 40 mTherefore, in 1s the distance covered by farmer = 40 / 40 m = 1mTherefore, in 140s distance covered by farmer = 1 � 140 m = 140 m. Now, number of rotation to cover 140 along the boundary= Total Distance / Perimeter= 140 m / 40 m = 3.5 roundThus, after 3.5 round farmer will at point C of the field.Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m NORTH east from initial position

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*18) The speed of a motor car is reduced from 54 km/hr to 36 km/hr in 15 second. Find the distance covered by the car during this time. How much farther will the car travel before coming to rest? Assume that the retardation is uniform throughout the motion. ​

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*18) The speed of a motor car is reduced from 54 km/hr to 36 km/hr in 15 second. Find the distance covered by the car during this time. How much farther will the car travel before coming to rest? Assume that the retardation is uniform throughout the motion.