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kirill115 [55]
3 years ago
9

Car A travels along a freeway at 110kmh-1 going east. Car B, travelling west at 120kmh-1 spins out of control and veers towards

Car A (across the median strip) at an angle 30° off course. a. What is the velocity of Car B relative to Car A?
Physics
1 answer:
AfilCa [17]3 years ago
5 0
To calculate the velocity relative to car A, you have to add the opposite of the velocity of car A to car B.

the velocity of car B relative to car A:
(let east and north directions be positive)
v = v (car B) - v(car A)
v(east-west direction) = -120 * cos 30° - 110
v(north - south direction) = -120 * sin 30° - 0
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Answer:

Approximately 2.1\; \rm km, assuming that g = -9.8\; \rm m \cdot s^{-2}.

Explanation:

Let t denote the time required for the package to reach the ground. Let h(\text{initial}) and h(\text{final}) denote the initial and final height of this package.

\displaystyle h(\text{final}) = \frac{1}{2}\, g\, t^2 + h(\text{initial}).

For this package:

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Solve for t, the time required for the package to reach the ground after being released.

\displaystyle t^{2} = \frac{2\, (h(\text{final}) - h(\text{initial}))}{g}.

\begin{aligned} t &= \sqrt{\frac{2\, (h(\text{final}) - h(\text{initial}))}{g} \\ &\approx \sqrt{\frac{2\times (0\; \rm m - 2500\; \rm m)}{(-9.8\; \rm m \cdot s^{-2})}} \approx 22.588\; \rm s\end{aligned}.

Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at 95\; \rm m \cdot s^{-1}.) From calculations above, the package would travel forward at that speed for about 22.588\; \rm s. That corresponds to approximately:95\; \rm m \cdot s^{-1} \times 22.588\; \rm s \approx 2.1 \times 10^{3}\; \rm m = 2.1\; \rm km.

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One should learn to trust the primary source more because it is the real work of the researcher.

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