Question

A particle starts from rest and moves with a constant acceleration of $2\text{ m/s}^2$. What is the distance traveled by the particle in the $5^{\text{th}}$ second of its motion?

• A. $25\text{ m}$
• B. $9\text{ m}$
• C. $10\text{ m}$
• D. $12.5\text{ m}$

Hint:

For a body starting from rest with constant acceleration, the distances covered in successive seconds follow Galileo's law of odd numbers:
$1 : 3 : 5 : 7 : 9 : \dots$
In the $1^{\text{st}}$ second, distance $= \frac{a}{2}(1) = 1\text{ m}$.
In the $5^{\text{th}}$ second, distance $= 9 \times 1\text{ m} = 9\text{ m}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the specific distance covered by a accelerating particle during its fifth single second of travel, given that it starts from rest.

Step 2: Key Formula or Approach:
The distance traveled by a body in the $n^{\text{th}}$ second of its motion under constant acceleration is given by:
\[ S_n = u + \frac{a}{2}(2n - 1) \]
where:
$u$ is the initial velocity.
$a$ is the constant acceleration.
$n$ is the specific second of interest.

Step 3: Detailed Explanation:

• The particle starts from rest, so the initial velocity $u = 0$.

• The constant acceleration is given as $a = 2\text{ m/s}^2$.

• We are looking for the distance traveled in the $5^{\text{th}}$ second, so $n = 5$.

• Substituting these values into the formula:
\[ S_5 = 0 + \frac{2}{2}\left(2(5) - 1\right) \]
\[ S_5 = 1 \times (10 - 1) \]
\[ S_5 = 9\text{ m} \]

• Alternatively, this can be calculated by subtracting the total distance covered in $4$ seconds from the total distance covered in $5$ seconds:
\[ S_{\text{total}}(t) = ut + \frac{1}{2}at^2 \]
\[ S(5) = \frac{1}{2}(2)(5)^2 = 25\text{ m} \]
\[ S(4) = \frac{1}{2}(2)(4)^2 = 16\text{ m} \]
\[ S_5 = S(5) - S(4) = 25 - 16 = 9\text{ m} \]

Step 4: Final Answer:
The distance traveled by the particle in the $5^{\text{th}}$ second is $9\text{ m}$.