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- Subject
- Engineering Mechanicsmechanical-engineering-mcqs › engineering-mechanics
- Published
- 25 Jan 2019
- Last updated
- 28 May 2026
Explanation
Using the formula for distance under constant acceleration starting from rest, s = 0.5 × a × t², we get s = 0.5 × 5 × (5)² = 0.5 × 5 × 25 = 62.5 meters.
More Engineering Mechanics MCQs
Practice related questions from the same subject.
- 1.Which of the following statements is incorrect?
- 2.Identify the false statement among the following regarding the center of gravity (C.G.):
- 3.What occurs when two couples lying in the same plane have moments that are equal in magnitude but opposite in direction?
- 4.What is the name of the frictional force that acts once an object is already in motion?
- 5.For a ladder placed on a frictionless surface and leaning against a vertical wall, in which direction does the frictional force act at the top contact point?
- 6.What is the term for the ratio between limiting friction and the normal force?
- 7.A framed structure is considered perfect when it has how many members in relation to the number of joints?
- 8.When analyzing stresses in frames using the method of sections, the frame is split by an imaginary cut that should not intersect more than how many members with unknown forces?
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