An object's motion is illustrated using the graph below. (NOT DRAWN TO SCALE)Determine the total displacement of the object given the following values:A : 18.9 ms-1B : -10 ms-1C : 9.4 sD : 16.3 s

Question

BlackTom AI · Accepted Answer

The velocity–time graph’s displacement is the area under the v(t) curve (positive area above the time axis minus negative area below it). Here we split the motion into two triangular areas: a large positive triangle spanning from t = 0 to t = D with peak velocity A, and a smaller negative triangle below the axis from t = C to t = D with depth |B|. The total displacement = (area of large positive triangle) − (area of small negative triangle).

1) Relevant concepts
- Displacement = integral of velocity over time = geometric area under v–t graph.
- Area of a triangle = 1/2 × base × height.
- Positive area (above time axis) contributes additive displacement; negative area (below axis) subtracts.

2) Step-by-step reasoning and computation
- Treat the positive region as a triangle with base = D (total time until the instantaneous change at D) and height = A.
  Area_positive = (1/2) × base × height = (1/2) × D × A.
- The negative region is a triangle from t = C to t = D with base = (D − C) and height = |B|.
  Area_negative = (1/2) × (D − C) × |B|.
- Net displacement = Area_positive − Area_negative.

3) Insert the given values and compute each intermediate step
- A = 18.9 m s−1, B = −10 m s−1 → |B| = 10 m s−1, C = 9.4 s, D = 16.3 s.
- Area_positive = 0.5 × A × D = 0.5 × 18.9 × 16.3 = 0.5 × 308.07 = 154.035 m.
- Area_negative = 0.5 × |B| × (D − C) = 0.5 × 10 × (16.3 − 9.4) = 0.5 × 10 × 6.9 = 34.5 m.
- Total displacement = 154.035 − 34.5 = 119.535 m ≈ 119.54 m.

4) Why this answer is correct
Using the area-under-curve method gives the net displacement as the algebraic sum of positive and negative triangular areas. Plugging the given magnitudes and time intervals yields 119.535 m, which rounds to 119.54 m.

5) Why the other listed numerical choices are incorrect
- 54.33 m corresponds to 0.5·A·C − 0.5·|B|·(D − C) (using C as the positive base), which assumes the positive area ends at C rather than extending to D — a different interpretation of the graph.
- 108.66 m and −39.67 m do not match the geometric area calculation with the given A, B, C, D values and therefore are inconsistent with the correct area subtraction shown above.

Conclusion: The area calculation produces a net displacement of 119.535 m, which rounds to 119.54 m, confirming the provided answer of 119.54 m.

An object's motion is illustrated using the graph below. (NOT DRAWN TO SCALE)Determine the total displacement of the object given the following values:A : 18.9 ms-1B : -10 ms-1C : 9.4 sD : 16.3 sNumerical

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