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Zolol [24]
2 years ago
14

In anticipation of a long 10o upgrade, a bus driver accelerates at a constant rate of 5 ft/s^2 while still on a level section of

a highway. Knowing that the speed of the bus was 80 mph as it begins to go up the hill and that the driver does not change the setting on his throttle or shift gears, determine the distance traveled (in miles) by the bus up the hill when its speed decreased to 50 mph.
Physics
1 answer:
Rashid [163]2 years ago
5 0

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, a_{Net} = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²

The vertical component of the velocity, v_y = v × sin(θ)

∴ v_y = 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 - a_{Net} × t

∴ t = (v₁ - v₂)/a_{Net}  = (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·a_{Net}·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

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Joe drove at the speed of 45 miles per hour for a certain distance. He then drove at the speed of 55 miles per hour for the same
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an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall, what is the height of
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The height of the object will be -5.19 cm

A concave mirror's reflecting surface curves inward and away from the light source. Light is reflected inward to a single focus point via concave mirrors. Concave mirrors, in contrast to convex mirrors, produce a variety of images depending on the object's to the mirror.

Given an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall

So let,

v =  Image distance from the mirror = -33.5 cm

u = object distance from the mirror (concave) = 24 cm

hi = Image height = 7.25 cm

h = height of the object = ?

Using below formula to find height of the object

-v/u = hi/h

Putting all value in the formula we get

-(-33.5)/(-24) = 7.25/h

h = -5.19 cm

Therefore the height of the object will be -5.19 cm

Learn more about Concave mirror here:

brainly.com/question/3727024

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