A car runs at a constant speed of 15ms-1 for 30secs, and then accelerates uniformly to a speed of 25ms-1 over a time of 20secs. this speed is maintain for the next 300secs before the car is brought to rest with uniform deceleration in 30secs. a. draw the velocity time graph of the motion. and calculate
b. the total distance traveled by the car
c. average speed of the car

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

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