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A car, initially at rest, moves along a straight road with constant acceleration such that it attains a velocity of 60 ft/s when s = 150 ft Then after being subjected to another constant acceleration it attains a final velocity of 100 ft/s when s = 325 ft. Determine the average velocity and average acceleration of the car for the entire 325-ft displacement.

A. V_avg = 55.0 ft/s, a_avg = 15.15 ft/s^2

B. V_avg = 45.2 ft/s, a_avg = 13.91 ft/s^2

C. V_avg = 80.0 ft/s, a_avg = 12.57 ft/s^2

D. V_avg = 80.0 ft/s, a_avg = 15.15 ft/s^2

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  1. 27 July, 22:45
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    B. V_avg = 45.2 ft/s, a_avg = 13.91 ft/s²

    Explanation:

    Given

    v₀ = 0 ft/s

    v₁ = 60 ft/s

    s₁ = 150 ft

    v₂ = 100 ft/s

    st = 325 ft

    Part 1

    We get a₁ as follows

    v₁² = v₀² + 2*a₁*s₁ ⇒ a₁ = (v₁² - v₀²) / (2*s₁)

    ⇒ a₁ = ((60 ft/s) ² - (0 ft/s) ²) / (2*150 ft) = 12 ft/s²

    then we obtain t₁

    v₁ = v₀ + a₁*t₁ ⇒ t₁ = (v₁ - v₀) / a₁

    ⇒ t₁ = (60 ft/s - 0 ft/s) / 12 ft/s² = 5 s

    Part 2

    We find s₂

    s₂ = st - s₁ ⇒ s₂ = 325 ft - 150 ft = 175 ft

    Now, we get a₂

    v₂² = v₁² + 2*a₂*s₂ ⇒ a₂ = (v₂² - v₁²) / (2*s₂)

    ⇒ a₂ = ((100 ft/s) ² - (60 ft/s) ²) / (2*175 ft) = 18.2857 ft/s²

    then we obtain t₂

    v₂ = v₁ + a₂*t₂ ⇒ t₂ = (v₂ - v₁) / a₂

    ⇒ t₂ = (100 ft/s - 60 ft/s) / 18.2857 ft/s² = 2.1875 s

    We find t = t₁ + t₂ = 5 s + 2.1875 s = 7.1875 s

    The average velocity will be

    v_avg = st/t = 325 ft/7.1875 s = 45.217 ft/s

    The average acceleration of the car for the entire 325-ft displacement will be

    a_avg = (v₂ - v₀) / t

    ⇒ a_avg = (100 ft/s - 0 ft/s) / 7.1875 s = 13.913 ft/s²
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