If a distance of 50, m is covered in 1, minut | vedclass

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Question

(A) The distances covered in consecutive minutes are as follows:Distance in $1^{st}$ minute $= 50\, m$.Distance in $2^{nd}$ minute $= 90 - 50 = 40\, m

Vedclass · Accepted Answer

$610$

Vedclass · Answer

(A) The distances covered in consecutive minutes are as follows:Distance in $1^{st}$ minute $= 50\, m$.Distance in $2^{nd}$ minute $= 90 - 50 = 40\, m$.Distance in $3^{rd}$ minute $= 130 - 90 = 40\, m$.Since the distance covered in each subsequent minute is constant $(40\, m)$,this forms an arithmetic progression where the first term $a = 50$ and the common difference $d = 40$.The total distance covered in $n$ minutes is given by the sum of the first $n$ terms of an arithmetic progression: $S_n = \frac{n}{2} [2a + (n - 1)d]$.For $n = 15$:$S_{15} = \frac{15}{2} [2(50) + (15 - 1)40]$$S_{15} = \frac{15}{2} [100 + 14 	imes 40]$$S_{15} = \frac{15}{2} [100 + 560]$$S_{15} = \frac{15}{2} 	imes 660 = 15 	imes 330 = 4950\, m$.Wait,re-evaluating the question: The values $50, 90, 130$ represent the *total* distance covered up to that time. Thus,the sequence is $50, 90, 130, \dots$ which is an arithmetic progression with $a = 50$ and $d = 40$.The distance at $n = 15$ is the $15^{th}$ term: $a_{15} = a + (n - 1)d$.$a_{15} = 50 + (15 - 1)40 = 50 + 14 	imes 40 = 50 + 560 = 610\, m$.

If a distance of $50\, m$ is covered in $1\, minute$,$90\, m$ in $2\, minutes$,and $130\, m$ in $3\, minutes$,find the total distance covered in $15\, minutes$ (in $m$).

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