An object's movement is depicted by the following motion graph. [NOT DRAWN TO SCALE]The particle travels a distance of 251.1 m during the motion Find the maximum velocity, v, of the particle given the following values.A : 5.1 sB : 17.2 sC : 26.5 s

Question

BlackTom AI · Accepted Answer

To find the maximum velocity from the velocity-time graph, I’ll break the motion into the three distinct portions shown: acceleration from rest to the top velocity, constant-velocity motion, and deceleration back to rest.

First, identify the time intervals:
- Acceleration phase ends at t = A = 5.1 s, reaching velocity V.
- Constant-velocity phase lasts from t = A to t = B, so duration is B − A = 17.2 − 5.1 = 12.1 s.
- Deceleration phase ends at t = C = 26.5 s, so duration of deceleration is C − B = 26.5 − 17.2 = 9.3 s.

Next, compute the total distance as the area under the velocity-time graph, which equals the sum of areas of three shapes:
- Area of the acceleration triangle: (1/2) × base (A) × height (V) = 0.5 × 5.1 × V = 2.55 V.
- Area of the constant-velocity rectangle: base (B − A) × height (V) = 12.1 × V = 12.1 V.
- Area of the deceleration triangle: (1/2) × base (C − B) × height (V) = 0.5 × 9.3 × V = 4.65 V.

Total distance traveled is the sum of these areas:
Distance = 2.55 V + 12.1 V + 4.65 V = (2.55 + 12.1 + 4.65) V = 19.3 V.

We’re given the distance as 251.1 m, so set 19.3 V = 251.1 and solve for V:
V = 251.1 / 19.3 ≈ 13.0 m/s (since 13 × 19.3 = 250.9, very close to the given distance).

Therefore, the maximum velocity is approximately 13.0 m/s.

Note on units and consistency: all altitudes and times are in seconds and meters, ensuring the area calculation yields meters, matching the distance given.

MUF0121 Physics Unit 1 - Semester 2, 2025

An object's movement is depicted by the following motion graph. [NOT DRAWN TO SCALE]The particle travels a distance of 251.1 m during the motion Find the maximum velocity, v, of the particle given the following values.A : 5.1 sB : 17.2 sC : 26.5 s

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