A body falls 16, m in the first second of its | vedclass

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(D) The distances covered by the body in the first,second,third,fourth,... seconds form an Arithmetic Progression $(A.P.)$: $16, 48, 80, 112, \dots$He

Vedclass · Accepted Answer

$336$

Vedclass · Answer

(D) The distances covered by the body in the first,second,third,fourth,... seconds form an Arithmetic Progression $(A.P.)$: $16, 48, 80, 112, \dots$Here,the first term $a = 16$ and the common difference $d = 48 - 16 = 32$.The distance covered in the $n^{th}$ second is given by the $n^{th}$ term of the $A.P.$: $T_n = a + (n - 1)d$.For the $11^{th}$ second,$n = 11$.$T_{11} = 16 + (11 - 1) 	imes 32$$T_{11} = 16 + 10 	imes 32$$T_{11} = 16 + 320 = 336\, m$.

$A$ body falls $16\, m$ in the first second of its motion,$48\, m$ in the second,$80\, m$ in the third,$112\, m$ in the fourth and so on. How far does it fall during the $11^{th}$ second of its motion? (in $m$)

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