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

Answer: (B)

$6.25 \, \text{m/s}^2$

Answer 2

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+ ( $\frac{a}{2}$)(2 n− 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+ $\frac{a}{2}$(2 n+ 3)

Set Up Equations for s nand s n+2:

From the given distances:

102.5 = u + $\frac{a}{2}$(2 n − 1)

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

Subtract the First Equation from the Second:

115.0 − 102.5 = (u + $\frac{a}{2}$(2 n + 3)) − (u + $\frac{a}{2}$(2 n − 1))

12.5 = $\frac{a}{2}$ × 4

12.5 = 2 a

a= $\frac{12.5}{2}$ = 6.25 m/s²

Conclusion:

The acceleration of the body is 6.25 m/s².