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Answer:
The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles
Explanation:
The parameters of the motion of the driver are;
The upgrade of the road, θ = 10°
The rate of constant acceleration of the bus driver = 5 ft./s²
The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s
The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s
The acceleration due to gravity, g ≈ 32.1740 ft./s²
Therefore, we have;
The acceleration due to gravity down the incline plane, gₓ = g·sinθ
∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²
The net acceleration of the bus, on the incline plane, = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²
The vertical component of the velocity, = v × sin(θ)
∴ = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s
vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s
The velocity of the car, v₂, on the inclined plane is given as follows;
v₂ = v₁ - × t
∴ t = (v₁ - v₂)/ = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s
The distance covered, 's', is given as follows;
s = v₁·t - 1/2··t²
∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.
The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles
Answer:
Explanation:
As we know that box is initially at rest
So we will have
now as it will be displaced from initial position to final position then final speed of the box is reached to
now we know by work energy theorem that work done by all the forces on the box will be equal to the change in kinetic energy
So here we will have