A ball rolling up an inclined plane covers 30 m during the first sec 27m during second sec and 24m during third second and so on. How much distance does the ball travel during the 10 sec and also find the total distance covered during 10 sec

Dear Student,

The distance covered in mts by ball in first sec, second sec, third sec, and so on is30, 27, 24, .....The above list is in AP, with first term, a = 30 and common difference, d = -3Now, distance covered in 10th second = a10 = a+9d = 30 + 9×-3 = 30 - 27 = 3 mSum of first n terms of an AP is,Sn = n2 2a + n-1dS10 = 1022×30 + 9×-3 = 560 - 27 = 5 × 33 = 165 So, distance covered in 10 sec = 165 m

Regards

  • 5
why should i tel.huh !!
  • -3
3m, 165m
  • -1
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