A particle moves with uniform acceleration along a straight line from rest. The percentage increase in displacement during the sixth’ second compared to that in the fifth second is about
22%
Understanding Particle Motion with Uniform Acceleration
The problem describes a particle that starts from rest and moves along a straight line with uniform acceleration. We need to find the percentage increase in the displacement covered during the sixth second compared to the displacement covered during the fifth second.
Formula for Displacement in the n-th Second
For a particle moving with uniform acceleration, the displacement covered in the n-th second is given by the formula:
\( s_n = u + \frac{a}{2}(2n-1) \)
Where:
- \( s_n \) is the displacement in the n-th second
- \( u \) is the initial velocity
- \( a \) is the uniform acceleration
- \( n \) is the time in seconds (e.g., 5th second, 6th second)
Calculating Displacement in the Fifth Second
The particle starts from rest, so the initial velocity \( u = 0 \). For the fifth second, \( n = 5 \). Using the formula:
\( s_5 = 0 + \frac{a}{2}(2 \times 5 - 1) \)
\( s_5 = \frac{a}{2}(10 - 1) \)
\( s_5 = \frac{a}{2}(9) \)
\( s_5 = \frac{9a}{2} \)
So, the displacement during the fifth second is \( \frac{9a}{2} \).
Calculating Displacement in the Sixth Second
For the sixth second, \( n = 6 \). Using the same formula with \( u = 0 \):
\( s_6 = 0 + \frac{a}{2}(2 \times 6 - 1) \)
\( s_6 = \frac{a}{2}(12 - 1) \)
\( s_6 = \frac{a}{2}(11) \)
\( s_6 = \frac{11a}{2} \)
So, the displacement during the sixth second is \( \frac{11a}{2} \).
Calculating the Percentage Increase in Displacement
To find the percentage increase in displacement during the sixth second compared to the fifth second, we first calculate the increase in displacement:
Increase = \( s_6 - s_5 \)
Increase = \( \frac{11a}{2} - \frac{9a}{2} \)
Increase = \( \frac{11a - 9a}{2} \)
Increase = \( \frac{2a}{2} \)
Increase = \( a \)
The percentage increase is calculated using the formula:
\( \text{Percentage Increase} = \frac{\text{Increase}}{\text{Original Value}} \times 100\% \)
Here, the original value is the displacement in the fifth second (\( s_5 \)).
\( \text{Percentage Increase} = \frac{a}{s_5} \times 100\% \)
Substitute the value of \( s_5 \):
\( \text{Percentage Increase} = \frac{a}{\frac{9a}{2}} \times 100\% \)
\( \text{Percentage Increase} = \frac{a}{1} \times \frac{2}{9a} \times 100\% \)
\( \text{Percentage Increase} = \frac{2}{9} \times 100\% \)
\( \text{Percentage Increase} = \frac{200}{9} \% \)
Calculating the value:
\( \frac{200}{9} \approx 22.22\% \)
This value is closest to 22%.
Comparing with Options
The calculated percentage increase is approximately 22.22%. Let's compare this with the given options:
| Option | Value |
|---|---|
| 1 | 11% |
| 2 | 22% |
| 3 | 33% |
| 4 | 44% |
The value 22.22% is closest to 22%.
Revision Table: Key Concepts in Uniformly Accelerated Motion
| Concept | Description | Formula (starting from rest, u=0) |
|---|---|---|
| Displacement (total) | Total distance covered in time \( t \) | \( s = ut + \frac{1}{2}at^2 \implies s = \frac{1}{2}at^2 \) |
| Velocity | Rate of change of displacement | \( v = u + at \implies v = at \) |
| Displacement in n-th second | Distance covered specifically during the n-th second | \( s_n = u + \frac{a}{2}(2n-1) \implies s_n = \frac{a}{2}(2n-1) \) |
| Uniform Acceleration | Constant rate of change of velocity | \( a = \text{constant} \) |
Additional Information: Understanding Motion in Specific Seconds
It is important to distinguish between the total displacement after a certain time \(t\) and the displacement during a specific second (like the 5th or 6th second). The displacement during the n-th second is the difference between the total displacement after \(n\) seconds and the total displacement after \(n-1\) seconds.
\( s_n = s(n) - s(n-1) \)
Using the formula \( s(t) = ut + \frac{1}{2}at^2 \), and with \( u = 0 \):
\( s(n) = \frac{1}{2}an^2 \)
\( s(n-1) = \frac{1}{2}a(n-1)^2 \)
So, \( s_n = \frac{1}{2}an^2 - \frac{1}{2}a(n-1)^2 \)
\( s_n = \frac{1}{2}a [n^2 - (n^2 - 2n + 1)] \)
\( s_n = \frac{1}{2}a [n^2 - n^2 + 2n - 1] \)
\( s_n = \frac{1}{2}a (2n - 1) \)
\( s_n = \frac{a}{2} (2n - 1) \)
This confirms the formula used for displacement in the n-th second when starting from rest. This method provides an alternative way to derive the formula and understand the concept of displacement in specific time intervals for uniformly accelerated motion.
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