Question

A body travels \( 102.5 \, \text{m} \) in \( n^\text{th} \) second and \( 115.0 \, \text{m} \) in \( (n+2)^\text{th} \) second. The acceleration is:

• A. \( 9 \, \text{m/s}^2 \)
• B. \( 6.25 \, \text{m/s}^2 \)
• C. \( 12.5 \, \text{m/s}^2 \)
• D. \( 5 \, \text{m/s}^2 \)

Answer 1

The Correct Option is B

Formula for Distance Travelled in the nth Second

The distance s_n travelled by a body in the nth second is given by:

s_n = 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: s_n = 102.5 m

Distance travelled in the (n + 2)^th second: s_n+2 = 115.0 m

Using the Formula for the (n + 2)th Second:

s_n+2 = u + $\frac{a}{2}$(2n + 3)

Set Up Equations for s_n and s_n+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.