Question 17: Motion in 1D

Question

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

Accepted Answer

The distance covered by a body in the $$n^{	ext{th}}$$ second can be found using the equation:
$$S_{n} = u + \dfrac{1}{2}a(2n-1)$$
where,
$$S_{n}$$ is the distance covered in the $$n^{	ext{th}}$$ second,
$$u$$ is the initial velocity,
$$a$$ is the acceleration, and
$$n$$ is the nth second.
The distance covered in the $$n^{	ext{th}}$$ second is given as $$102.5 \, 	ext{m}$$, so we have:
$$102.5 = u + \dfrac{1}{2}a(2n-1)$$ ---- (1)
For the $$(n + 2)^{	ext{th}}$$ second, the distance covered is:
$$S_{n+2} = u + \dfrac{1}{2}a(2(n+2)-1)$$
Substituting $$n + 2$$ in place of $$n$$, we get:
$$115.0 = u + \dfrac{1}{2}a(2n+3)$$ ---- (2)
Subtracting equation (1) from equation (2), we get:
$$115.0 - 102.5 = \dfrac{1}{2}a(2n + 3 - 2n + 1)$$
$$12.5 = \dfrac{1}{2}a(4)$$
So, solving for $$a$$ gives:
$$a = \dfrac{12.5 	imes 2}{4} = \dfrac{25}{4} = 6.25 \, 	ext{m/s}^2$$
Therefore, the acceleration of the body is:
$$6.25 \, 	ext{m/s}^2$$
Which corresponds to Option A: $$6.25 \, 	ext{m/s}^2$$.

Answer

$$6.25 \mathrm{~m} / \mathrm{s}^2$$

Answer

$$5 \mathrm{~m} / \mathrm{s}^2$$

Answer

$$12.5 \mathrm{~m} / \mathrm{s}^2$$

Answer

$$9 \mathrm{~m} / \mathrm{s}^2$$

Motion in 1D

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

Choose the correct answer: