Question

*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 1

Answer:

Here, Side of the given square field = 10m

Here, Side of the given square field = 10mso, perimeter of a square = 4*side = 10 m * 4 = 40 m

Here, Side of the given square field = 10mso, perimeter of a square = 4*side = 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 = 4*side = 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 seconds

Here, Side of the given square field = 10mso, perimeter of a square = 4*side = 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 m

Here, Side of the given square field = 10mso, perimeter of a square = 4*side = 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 = 1m

Here, Side of the given square field = 10mso, perimeter of a square = 4*side = 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

Here, Side of the given square field = 10mso, perimeter of a square = 4*side = 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 round

Here, Side of the given square field = 10mso, perimeter of a square = 4*side = 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 = 4*side = 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

Answer 2

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