Trigonometric components and motion models
導入済み| English | 意味 |
|---|---|
| component/kəmˈpəʊnənt/ | 成分 |
A speed, force and height can each use sine or cosine. Which reference angle and axis decides the formula?
- A speed, force and height can each use sine or cosine. Which reference angle and axis decides the formula?
- This lesson studies component 分量: The signed projection of a vector along a chosen coordinate direction.
Choose the mathematical structure
- A vector of magnitude R at angle θ anticlockwise from the positive horizontal axis has components (Rcosθ,Rsinθ). A bearing β measured clockwise from north instead gives east Rsinβ and north Rcosβ. Preserve units and signs. For a periodic height c+a sin(ωt+φ), a full cycle has duration 2π/|ω| when ω≠0; interpret the time interval and physical assumptions.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
A velocity of 12 m/s is at bearing 060°. Why is the north component 6, not 6√3?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Bearings turn clockwise from north, swapping the usual roles: east=R sin β and north=R cos β. Using sine for north would assign the 30°-from-east value to the wrong axis.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
A velocity of 12 metres per second at bearing 60° has east component 6√3 and north component 6 metres per second; its horizontal direction angle is 30°. For a 2 kg block on an ideal frictionless 30° slope, take g=10 metres per second squared: weight=20 N, downslope component=20sin30°=10 N and normal component=20cos30°=10√3 N. If there are no other normal forces, the reaction has that normal magnitude. For a uniformly rotating wheel, centre height 3 m and radius 2 m give h(t)=3+2sin(πt/6), with t in seconds and phase zero at t=0. Its period is 12 s and range 1–5 m. On 0≤t<12, height 4 m occurs at t=1 and 5 s; the first is 1 s. Differentiation gives vertical velocity (π/3)cos(πt/6), zero at the highest point t=3 s.
Trigonometric components and motion models
A vector of magnitude R at angle θ anticlockwise from the positive horizontal axis has components (Rcosθ,Rsinθ)
Explain the original interval, denominator or physical reference before using a trig equation.
For speed 12 at bearing 60°, find the north component in metres per second.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
The north axis is adjacent to the bearing: 12cos60°=6.
Test a tempting shortcut
- A bearing is not an angle from the horizontal. The downslope component of weight uses sine of the slope angle; the normal uses cosine. A reaction does not always equal the weight’s normal component if other forces act normally. A position or height function is not itself a velocity. Negative components show direction, not negative magnitude.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
For every bearing, the north component is the magnitude multiplied by the sine of the bearing. This claim is false. Explain which definition or assumption it violates.
For the stated slope model, find the downslope weight component in newtons.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Weight=2×10=20 N, then downslope=20sin30°=10 N.
For every bearing, the north component is the magnitude multiplied by the sine of the bearing.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
A bearing is not an angle from the horizontal. The downslope component of weight uses sine of the slope angle; the normal uses cosine. A reaction does not always equal the weight’s normal component if other forces act normally. A position or height function is not itself a velocity. Negative components show direction, not negative magnitude.
Interpret a new situation
- Draw and label axes, angle origin and the interval before projecting or solving. Recover magnitude with √(u²+v²) and choose the direction quadrant from both signed components. State the ideal assumptions: fixed rotation rate for the wheel; no friction or extra normal forces for the block. Restrict mathematical roots to physically allowed times and check dimensions.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the first nonnegative time when the wheel height is 4 m, in seconds.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
πt/6=π/6 first gives t=1 second; the other cycle solution is 5 seconds.
Match each periodic-model quantity for h=3+2sin(πt/6) to its value.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
In c+a sin(ωt+φ), the centre line, amplitude and period read off directly; the model is interpreted only over its stated time interval.
Use this in your course
- 7357 · A-level · E. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The signed projection of a vector along a chosen coordinate direction. Choose the relationship, show the method, check its assumptions and interpret the result.