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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jolis1796 jolis1796 · Answer 1 · 2020-02-12T06:29:38+01:00

Answer:

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