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In a potato race, a bucket is placed at the starting point, which is $5$ m from the first potato. The other potatoes are arranged $3$ m apart in a straight line, with a total of $10$ potatoes, as shown in the figure :
A competitor starts from the bucket, picks up the nearest potato, runs back to the bucket to drop it in, then returns to pick up the next potato. This process continues until all the potatoes are in the bucket.
Based on the above information, answer the following questions :
(i) What is the distance covered to pick up the first potato and drop it in bucket?
(ii) What is the distance covered to pick up the second potato and drop it in bucket ?
(iii) (a) What is the total distance the competitor has to run ?
OR
(iii) (b) If average speed of competitor is $5$ m/s, then find the average time taken by competitor to put all the potatoes in the bucket.
A competitor starts from the bucket, picks up the nearest potato, runs back to the bucket to drop it in, then returns to pick up the next potato. This process continues until all the potatoes are in the bucket.
Based on the above information, answer the following questions :
(i) What is the distance covered to pick up the first potato and drop it in bucket?
(ii) What is the distance covered to pick up the second potato and drop it in bucket ?
(iii) (a) What is the total distance the competitor has to run ?
OR
(iii) (b) If average speed of competitor is $5$ m/s, then find the average time taken by competitor to put all the potatoes in the bucket.
Show SolutionHide Solution↓
(i) Required distance for the first potato $= 5 + 5 = 10$ m (I) (1 Mark)
(ii) Required distance for the second potato $= 8 + 8 = 16$ m (I) (1 Mark)
(iii) (a) Distances covered form an A.P. with $a = 10$ and $d = 6$ (I) (1 Mark)
$\therefore S_{10} = \frac{10}{2} [2 \times 10 + 9 \times 6]$
$= 5 \times 74 = 370$ m (II) (1 Mark)
OR
(iii) (b) Distances covered form an A.P. with $a = 10$ and $d = 6$ (I) (1/2 Mark)
$\therefore S_{10} = \frac{10}{2} [2 \times 10 + 9 \times 6]$
$= 5 \times 74 = 370$ m (II) (1 Mark)
Time $= \frac{370}{5} = 74$ seconds (III) (1/2 Mark)
(ii) Required distance for the second potato $= 8 + 8 = 16$ m (I) (1 Mark)
(iii) (a) Distances covered form an A.P. with $a = 10$ and $d = 6$ (I) (1 Mark)
$\therefore S_{10} = \frac{10}{2} [2 \times 10 + 9 \times 6]$
$= 5 \times 74 = 370$ m (II) (1 Mark)
OR
(iii) (b) Distances covered form an A.P. with $a = 10$ and $d = 6$ (I) (1/2 Mark)
$\therefore S_{10} = \frac{10}{2} [2 \times 10 + 9 \times 6]$
$= 5 \times 74 = 370$ m (II) (1 Mark)
Time $= \frac{370}{5} = 74$ seconds (III) (1/2 Mark)