Mechanics
⏱ ~3-min readAceMark GuideWhat this topic is really about
The horizontal range of a projectile is maximized when the term sin(2θ) in the range formula reaches its maximum value of 1, which occurs at 2θ = 90 degrees, or θ = 45 degrees. Launching at 90 degrees (Option D) is incorrect because it results in purely vertical motion and zero horizontal range, while 30 degrees (Option A) yields a shorter range.
The coefficient of friction, μ, is defined as the ratio of the limiting frictional force to the normal reaction force between two contacting surfaces, making B correct. Option C is incorrect because mass and gravity determine the weight of an object, not the direct ratio of friction to reaction, though weight can influence the normal reaction.
See the mechanism
Differentiating the displacement equation gives the velocity function v = 8t - 3. A diagram for this topic isn't available yet — the worked example below walks the same reasoning step by step.
An exam-style question, fully explained
A particle moves with s = 4t² − 3t. At t = 2, velocity ds/dt =
- Identify what the question tests: A particle moves with s = 4t² − 3t..
- Differentiating the displacement equation gives the velocity function v = 8t - 3.
- Substituting t = 2 into this function yields a velocity of 13, which makes B correct.
- Option A is incorrect because it is the result of an arithmetic error, while C and D fail to apply the differentiation rules correctly.
Traps the examiner sets
- Option A is incorrect because it is the result of an arithmetic error, while C and D fail to apply the differentiation rules correctly.
- Launching at 90 degrees (Option D) is incorrect because it results in purely vertical motion and zero horizontal range, while 30 degrees (Option A) yields a shorter range.
- Option C is incorrect because mass and gravity determine the weight of an object, not the direct ratio of friction to reaction, though weight can influence the normal reaction.
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