A race car moving along a circular track has a centripetal acceleration of 15.4 m/s? If the car has
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Answer:
a. Please find the attached velocity time graph of the car's motion created with Microsoft Excel
b. The total distance traveled by the car is 8,725 meters
c. The average speed of the car is 22.9605263 m/s
Explanation:
The given parameters of the motion are;
The initial speed of the car, v₁ = 15 m/s
The time during which the car runs at the initial speed, t₁ = 30 seconds
The new speed the car then accelerates at 'a₁' to, v₂ = 25 m/s
The duration it takes for the car to accelerate to the new speed = 20 seconds
The time during which the car runs at the initial speed = 300 seconds
The time it takes the car to be brought to rest with a deceleration, 'a₂' from the new speed (20 m/s) = 30 seconds
The final speed of the car at rest, v₃ = 0 m/s
The acceleration, a₁ = (v₂ - v₁)/t₁ = (25 - 15)/20 = 1/2 m/s²
The deceleration , a₂ = (v₃ - v₂)/t₁ = (0 - 25)/30 = -5/6 m/s²
a. Please find attached the drawing of the velocity time graph of the motion created with Microsoft Excel
b. The total distance traveled by the car, 'Δx', is given b the area under the velocity time graph as follows;
Area of trapezoid, A₁ = (320 + 300)/2 × 10 = 3,100
Area of rectangle, A₂ = 15 × 350 = 5,250
Area of triangle, A₃ = 1/2×30×25 = 375
The total area under the velocity time graph = A₁ + A₂ + A₃ = 3,100 + 5,250 + 375 = 8,725
The total area under the velocity time graph = The total distance traveled by the car, Δx = 8,725 meters
c. The average speed of the car is given as follows;
Where;
Δt = The total time during which the car travels
∴ The average speed of the car = 8,725 m/(380 s) = 22.9605263 m/s
