A ball rolling up an incline covers 36, m dur | vedclass

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Question

(A) The distances covered by the ball in successive seconds are $36\, m, 32\, m, 28\, m, \dots$This sequence forms an Arithmetic Progression $(A.P.)$

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The distances covered by the ball in successive seconds are $36\, m, 32\, m, 28\, m, \dots$This sequence forms an Arithmetic Progression $(A.P.)$ where the first term $a = 36$ and the common difference $d = 32 - 36 = -4$.The distance covered during the $n^{th}$ second is given by the formula for the $n^{th}$ term of an $A.P.$:$T_n = a + (n - 1)d$For the $8^{th}$ second,$n = 8$:$T_8 = 36 + (8 - 1)(-4)$$T_8 = 36 + 7(-4)$$T_8 = 36 - 28$$T_8 = 8\, m$Therefore,the ball will travel $8\, m$ during the $8^{th}$ second.

$A$ ball rolling up an incline covers $36\, m$ during the first second,$32\, m$ during the second,$28\, m$ during the next and so on. How much distance will it travel during the $8^{th}$ second? (in $m$)

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