The distance sn travelled by a body in the nth second is given by:
sn = u + (\(\frac{a}{2}\))(2n − 1)
where u is the initial velocity, a is the acceleration, and n is the specific second.
Given Data: Distance travelled in the nth second: sn = 102.5 m
Distance travelled in the (n + 2)th second: sn+2 = 115.0 m
Using the Formula for the (n + 2)th Second:
sn+2 = u + $\frac{a}{2}$(2n + 3)
Set Up Equations for sn and sn+2:
From the given distances:
102.5 = u + $\frac{a}{2}$(2n − 1)
115.0 = u + $\frac{a}{2}$(2n + 3)
Subtract the First Equation from the Second:
115.0 − 102.5 = (u + $\frac{a}{2}$(2n + 3)) − (u + $\frac{a}{2}$(2n − 1))
12.5 = $\frac{a}{2}$ × 4
12.5 = 2a
a = $\frac{12.5}{2}$ =6.25 m/s²
Conclusion:
The acceleration of the body is 6.25 m/s².
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Approach Solution -2
Step 1: Given data.
The distance travelled in the \( n^{th} \) second, \( s_n = 102.5 \, \text{m} \)
The distance travelled in the \( (n+2)^{th} \) second, \( s_{n+2} = 115.0 \, \text{m} \).
We are asked to find the acceleration \( a \).
Step 2: Formula for distance travelled in the \( n^{th} \) second.
The distance travelled in the \( n^{th} \) second under uniform acceleration is given by:
\[
s_n = u + \frac{a}{2}(2n - 1)
\]
where \( u \) is the initial velocity and \( a \) is the acceleration.
Step 3: Write equations for both cases.
For the \( n^{th} \) second:
\[
s_n = u + \frac{a}{2}(2n - 1) = 102.5
\]
For the \( (n + 2)^{th} \) second:
\[
s_{n+2} = u + \frac{a}{2}[2(n + 2) - 1] = u + \frac{a}{2}(2n + 3) = 115.0
\]
Step 4: Subtract the two equations.
\[
s_{n+2} - s_n = \frac{a}{2}[(2n + 3) - (2n - 1)] = \frac{a}{2}(4) = 2a
\]
\[
115.0 - 102.5 = 12.5 = 2a
\]
\[
\Rightarrow a = 6.25 \, \text{m/s}^2
\]
Step 5: Final Answer.
\[
\boxed{a = 6.25 \, \text{m/s}^2}
\]
