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. 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.
(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 Hint
Always remember to double the distance in "back and forth" problems; the most common mistake is calculating the distance to the object only.
Step 1: Understanding the Concept:
The distances covered for each potato form an Arithmetic Progression (AP). Since the competitor runs to the potato and back to the bucket, the distance for each potato is twice the distance of the potato from the bucket. Step 2: Key Formula or Approach:
1. Distance for $n^{th}$ potato: $2 \times [\text{Distance from bucket}]$.
2. Sum of AP: $S_n = \frac{n}{2}[2a + (n-1)d]$. Step 3: Detailed Explanation:
1. (i) First potato: Distance = $2 \times 5 = 10$ m.
2. (ii) Second potato: Distance = $2 \times (5 + 3) = 2 \times 8 = 16$ m.
3. (iii) (a) Total distance:
- The sequence is $10, 16, 22, \dots$ up to 10 terms.
- $a = 10, d = 6, n = 10$.
- $S_{10} = \frac{10}{2}[2(10) + (10-1)6] = 5[20 + 54] = 5 \times 74 = 370$ m.
4. (iii) (b) OR Time taken:
- Total distance = $370$ m.
- $\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{370}{5} = 74$ seconds. Step 4: Final Answer:
(i) 10 m. (ii) 16 m. (iii)(a) 370 m or (iii)(b) 74 seconds.
