If a body moving with uniform acceleration travels a distance of 10 m in the first 10 s, then how much total distance in m will it cover at the end of 30 s from the beginning?

Question

A body starts from rest and moves with constant acceleration. If it covers a distance of $ 40 \, \text{m} $ in 4 seconds, then its acceleration is:

Answer 1

Step 1: Starting from rest, distance covered is proportional to the square of time: $ s \propto t^2 $, so $ \dfrac{s_{30}}{s_{10}} = \left(\dfrac{30}{10}\right)^2 = 9 $.
Step 2: Since $ s_{10} = 10 \, \text{m} $, the distance at 30 s is $ s_{30} = 9 \times 10 = 90 \, \text{m} $.
Step 3: This proportional-scaling shortcut avoids solving for the acceleration explicitly and gives the same result directly.
\[ \boxed{s_{30} = 90 \, \text{m}} \]

Answer 2

Another way to see this is through the velocity-time graph. Since the body starts from rest with uniform acceleration, its velocity-time graph is a straight line through the origin, and the distance covered equals the area under this line (a triangle). First, find the acceleration from the first 10 s: $ 10 = \frac{1}{2}a(10)^2 \Rightarrow a = 0.2 \, \text{m/s}^2 $. At $ t = 30 \, \text{s} $, the velocity is $ v = at = 0.2 \times 30 = 6 \, \text{m/s} $. The area of the velocity-time triangle from 0 to 30 s is $ \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 30 \times 6 = 90 \, \text{m} $. Let's check each option against this graphical area.

30: This is far smaller than the triangular area computed (90 m) and does not match the geometry of the velocity-time graph over the full 30 s.
50: This is also smaller than the full triangle's area and would only correspond to a much shorter time interval, not the full 30 s.
70: This is close to but still short of the correct triangular area of 90 m, so it does not match the graph's area calculation.
90: This exactly matches the area of the velocity-time triangle from 0 to 30 s.

The area under the velocity-time graph confirms the total distance covered in 30 s is 90 m.

Therefore, the correct answer is 90 m.

Answer 3

Another way to see this is through the velocity-time graph. Since the body starts from rest with uniform acceleration, its velocity-time graph is a straight line through the origin, and the distance covered equals the area under this line (a triangle). First, find the acceleration from the first 10 s: $ 10 = \frac{1}{2}a(10)^2 \Rightarrow a = 0.2 \, \text{m/s}^2 $. At $ t = 30 \, \text{s} $, the velocity is $ v = at = 0.2 \times 30 = 6 \, \text{m/s} $. The area of the velocity-time triangle from 0 to 30 s is $ \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 30 \times 6 = 90 \, \text{m} $. Let's check each option against this graphical area.

30: This is far smaller than the triangular area computed (90 m) and does not match the geometry of the velocity-time graph over the full 30 s.
50: This is also smaller than the full triangle's area and would only correspond to a much shorter time interval, not the full 30 s.
70: This is close to but still short of the correct triangular area of 90 m, so it does not match the graph's area calculation.
90: This exactly matches the area of the velocity-time triangle from 0 to 30 s.

The area under the velocity-time graph confirms the total distance covered in 30 s is 90 m.

Therefore, the correct answer is 90 m.

If a body moving with uniform acceleration travels a distance of 10 m in the first 10 s, then how much total distance in m will it cover at the end of 30 s from the beginning?

Show Hint

The Correct Option is D

Approach Solution - 1

Approach Solution -2

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