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Calculate Constant Acceleration of a Cyclist with Speed Change Over 2 km Distance
Physics
Grade 10 of Junior High School
Question Content
A cyclist travels 2 km on a flat horizontal road, increasing their speed from 28 kmh⁻¹ to 32 kmh⁻¹. Assuming constant acceleration, find their acceleration in this period.
Correct Answer
0.0104 m/s² (or 0.01 m/s² when rounded to two decimal places)
Detailed Solution Steps
1
Step 1: Convert all units to the SI system (meters and seconds). Distance s = 2 km = 2000 m; initial speed u = 28 km/h = 28 * (1000/3600) ≈ 7.778 m/s; final speed v = 32 km/h = 32 * (1000/3600) ≈ 8.889 m/s.
2
Step 2: Select the appropriate kinematic equation for constant acceleration without time: v² = u² + 2as, where v is final velocity, u is initial velocity, a is acceleration, and s is displacement.
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Step 3: Rearrange the equation to solve for acceleration a: a = (v² - u²) / (2s).
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Step 4: Substitute the values into the formula: a = ((8.889)² - (7.778)²) / (2*2000) = (79.012 - 60.500) / 4000 = 18.512 / 4000 ≈ 0.0104 m/s².
Knowledge Points Involved
1
Unit Conversion for Kinematic Quantities
Kinematic calculations in physics typically use the SI system (meters for distance, seconds for time, m/s for velocity, m/s² for acceleration). To convert km/h to m/s, multiply by 1000/3600 (or 5/18). This ensures consistency in equations and valid results.
2
Kinematic Equations for Constant Acceleration
The equation v² = u² + 2as is one of the core kinematic equations for motion with constant acceleration. It relates initial velocity (u), final velocity (v), acceleration (a), and displacement (s), and is used when the time of motion is unknown or not required for the calculation.
3
Displacement and Velocity Relationship
Displacement refers to the straight-line distance between the start and end points of motion (here, the 2 km road length). For motion with constant acceleration, the change in velocity squared is directly proportional to displacement and acceleration, as shown in the selected kinematic equation.
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