To solve the problem step by step, we will analyze the motion of both persons A and B.
### Step 1: Analyze A's Motion
Person A starts from rest and accelerates along the x-axis with an acceleration of \(3.2 \, \text{m/s}^2\).
Using the equation of motion:
\[
s_1 = ut + \frac{1}{2} a t^2
\]
where \(u = 0\) (initial velocity), \(a = 3.2 \, \text{m/s}^2\), and \(s_1\) is the distance traveled during acceleration.
### Step 2: Find the Distance Traveled by A During Acceleration
Let \(t_1\) be the time taken to reach the maximum distance \(s_1\). We know that A comes to rest after traveling a total distance of \(320 \, \text{m}\), which includes both acceleration and deceleration phases.
Let \(s_2\) be the distance traveled during deceleration. Since A comes to rest, we can use the equation:
\[
s_2 = \frac{v^2 - u^2}{2a}
\]
where \(v = 0\) (final velocity), \(u\) is the velocity at the end of the acceleration phase, and \(a = -3.2 \, \text{m/s}^2\) (retardation).
### Step 3: Relate Distances
Since \(s_1 + s_2 = 320 \, \text{m}\), we can express \(s_2\) in terms of \(s_1\):
\[
s_2 = 320 - s_1
\]
### Step 4: Find the Final Velocity After Acceleration
The final velocity \(u\) at the end of the acceleration phase can be found using:
\[
u = at_1 = 3.2 t_1
\]
Now substituting this into the equation for \(s_2\):
\[
s_2 = \frac{(3.2 t_1)^2}{2 \cdot 3.2} = \frac{3.2 t_1^2}{2}
\]
### Step 5: Set Up the Equation
Now we have:
\[
s_1 + \frac{3.2 t_1^2}{2} = 320
\]
Using the equation for \(s_1\):
\[
s_1 = \frac{1}{2} \cdot 3.2 \cdot t_1^2
\]
Substituting this into the equation gives:
\[
\frac{1}{2} \cdot 3.2 \cdot t_1^2 + \frac{3.2 t_1^2}{2} = 320
\]
This simplifies to:
\[
3.2 t_1^2 = 320 \implies t_1^2 = 100 \implies t_1 = 10 \, \text{s}
\]
### Step 6: Find the Distance Traveled by A
Now substituting \(t_1\) back to find \(s_1\):
\[
s_1 = \frac{1}{2} \cdot 3.2 \cdot (10)^2 = 160 \, \text{m}
\]
Thus, \(s_2 = 320 - 160 = 160 \, \text{m}\).
### Step 7: Calculate the Velocity at the End of 10 seconds
The velocity at the end of the acceleration phase (after 10 seconds) is:
\[
v = 3.2 \cdot 10 = 32 \, \text{m/s}
\]
### Step 8: Calculate Distance Traveled in the Next 5 Seconds
During the next 5 seconds, A decelerates to rest:
\[
s_2 = v \cdot t - \frac{1}{2} a t^2
\]
Substituting \(v = 32 \, \text{m/s}\) and \(t = 5 \, \text{s}\):
\[
s_2 = 32 \cdot 5 - \frac{1}{2} \cdot 3.2 \cdot (5)^2 = 160 - 40 = 120 \, \text{m}
\]
### Step 9: Total Distance Traveled by A
The total distance traveled by A after 15 seconds:
\[
d_A = s_1 + s_2 = 160 + 120 = 280 \, \text{m}
\]
### Step 10: Calculate Distance Between A and B
The initial distance between A and B is \(210 \, \text{m}\). The distance B needs to cover to catch A is:
\[
d_{AB} = \sqrt{(280)^2 + (210)^2}
\]
Calculating:
\[
d_{AB} = \sqrt{78400 + 44100} = \sqrt{122500} = 350 \, \text{m}
\]
### Step 11: Find the Velocity of B
B catches A in \(15 \, \text{s}\):
\[
v_B = \frac{d_{AB}}{t} = \frac{350}{15} = \frac{70}{3} \, \text{m/s}
\]
### Step 12: Relate B's Speed to Given Expression
We know from the problem statement:
\[
v_B = \frac{10n}{3}
\]
Setting the two expressions equal:
\[
\frac{70}{3} = \frac{10n}{3}
\]
Thus, solving for \(n\):
\[
70 = 10n \implies n = 7
\]
### Final Answer
The value of \(n\) is \(7\).
