To solve the problem step by step, we will use the equations of motion for uniformly accelerated motion.
### Step 1: Understand the given data
- The body covers 200 m in the first 2 seconds.
- The body covers 220 m in the next 4 seconds (total time = 6 seconds).
- We need to find the velocity after 7 seconds.
### Step 2: Use the first equation of motion for the first segment
For the first 2 seconds, we can use the second equation of motion:
\[
s = ut + \frac{1}{2} a t^2
\]
Here, \(s = 200 \, \text{m}\), \(t = 2 \, \text{s}\), and we need to find \(u\) (initial velocity) and \(a\) (acceleration).
Substituting the values into the equation:
\[
200 = u \cdot 2 + \frac{1}{2} a \cdot (2^2)
\]
This simplifies to:
\[
200 = 2u + 2a
\]
Dividing the entire equation by 2:
\[
100 = u + a \quad \text{(Equation 1)}
\]
### Step 3: Use the second equation of motion for the total distance in 6 seconds
For the total distance covered in 6 seconds (200 m + 220 m = 420 m):
\[
s = ut + \frac{1}{2} a t^2
\]
Here, \(s = 420 \, \text{m}\), \(t = 6 \, \text{s}\).
Substituting the values:
\[
420 = u \cdot 6 + \frac{1}{2} a \cdot (6^2)
\]
This simplifies to:
\[
420 = 6u + 18a
\]
Dividing the entire equation by 6:
\[
70 = u + 3a \quad \text{(Equation 2)}
\]
### Step 4: Solve the system of equations
Now we have two equations:
1. \(100 = u + a\)
2. \(70 = u + 3a\)
We can subtract Equation 1 from Equation 2:
\[
(70 - 100) = (u + 3a) - (u + a)
\]
This simplifies to:
\[
-30 = 2a
\]
Thus, we find:
\[
a = -15 \, \text{m/s}^2
\]
### Step 5: Substitute \(a\) back to find \(u\)
Using the value of \(a\) in Equation 1:
\[
100 = u - 15
\]
So,
\[
u = 100 + 15 = 115 \, \text{m/s}
\]
### Step 6: Find the final velocity after 7 seconds
Using the first equation of motion:
\[
v = u + at
\]
Substituting \(u = 115 \, \text{m/s}\), \(a = -15 \, \text{m/s}^2\), and \(t = 7 \, \text{s}\):
\[
v = 115 + (-15) \cdot 7
\]
Calculating:
\[
v = 115 - 105 = 10 \, \text{m/s}
\]
### Final Answer
The velocity after 7 seconds is \(10 \, \text{m/s}\).
