Jack and Jill are separated by a distance of 96 𝑚𝑒𝑡𝑒𝑟𝑠 and start moving towards each other from
rest and accelerate constantly at 1.4 𝑚/𝑠
2
 and 1.5 𝑚/𝑠
2
, respectively. A bee, starting from Jack’s head
from rest and with a constant acceleration of 1.6 𝑚/𝑠
2
starts flying towards Jill, reaches her, flies back to
Jack and continues between them until they meet. How much distance did the bee cover? 

We can approach this problem step by step by first calculating when Jack and Jill meet and then how much distance the bee covers during that time.

### Step 1: Time until Jack and Jill meet

Jack's acceleration: \( a_J = 1.4 \, \text{m/s}^2 \)  
Jill's acceleration: \( a_j = 1.5 \, \text{m/s}^2 \)  
Distance between Jack and Jill: \( d = 96 \, \text{m} \)

The distance covered by Jack is given by:
\[
d_J = \frac{1}{2} a_J t^2
\]
The distance covered by Jill is:
\[
d_j = \frac{1}{2} a_j t^2
\]
Since Jack and Jill are moving towards each other, their total distance covered together equals 96 meters:
\[
d_J + d_j = 96
\]
Substituting the expressions for \( d_J \) and \( d_j \):
\[
\frac{1}{2} a_J t^2 + \frac{1}{2} a_j t^2 = 96
\]
Factoring out \( \frac{1}{2} t^2 \):
\[
\frac{1}{2} t^2 (a_J + a_j) = 96
\]
Now, plug in the values for \( a_J = 1.4 \, \text{m/s}^2 \) and \( a_j = 1.5 \, \text{m/s}^2 \):
\[
\frac{1}{2} t^2 (1.4 + 1.5) = 96
\]
\[
\frac{1}{2} t^2 (2.9) = 96
\]
\[
t^2 = \frac{96 \times 2}{2.9}
\]
\[
t^2 = \frac{192}{2.9}
\]
\[
t^2 \approx 66.21
\]
\[
t \approx \sqrt{66.21} \approx 8.14 \, \text{seconds}
\]

### Step 2: Distance covered by the bee

The bee starts with an acceleration of \( a_B = 1.6 \, \text{m/s}^2 \). The total time the bee is flying is the same time it takes Jack and Jill to meet, which is approximately \( t = 8.14 \, \text{seconds} \).

The distance covered by the bee can be calculated using the formula for uniformly accelerated motion:
\[
d_B = \frac{1}{2} a_B t^2
\]
Substitute the values for \( a_B = 1.6 \, \text{m/s}^2 \) and \( t = 8.14 \, \text{seconds} \):
\[
d_B = \frac{1}{2} \times 1.6 \times (8.14)^2
\]
\[
d_B = 0.8 \times 66.21
\]
\[
d_B \approx 52.97 \, \text{meters}
\]

### Final Answer:
The bee covers approximately **52.97 meters** before Jack and Jill meet.

Would you like more details or clarifications?

Here are 5 related questions for further exploration:
1. How would the result change if Jack or Jill's acceleration was higher or lower?
2. What if the bee had no acceleration but constant speed instead?
3. Can you calculate how far Jack and Jill traveled individually before meeting?
4. How does the bee’s flight path change if the accelerations of Jack, Jill, and the bee are time-dependent?
5. What would be the impact if there were air resistance acting on the bee?

**Tip:** Always ensure to write down the known values clearly to avoid mistakes when solving physics problems.

Jack and Jill are separated by a distance of 96 meters and start moving towards each other from rest and accelerate constantly at 1.4 m/s² and 1.5 m/s², respectively. A bee, starting from Jack’s head from rest and with a constant acceleration of 1.6 m/s², starts flying towards Jill, reaches her, flies back to Jack and continues between them until they meet. How much distance did the bee cover?

Solve a classic kinematic problem where Jack and Jill move towards each other, and a bee flies back and forth between them with constant acceleration. This problem involves calculating time and distance using kinematic equations for uniformly accelerated motion. Suitable for high school physics students.

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