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Kazeer [188]
3 years ago
10

A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exe

rts a resistive force with magnitude proportional to the square of the speed with k
Physics
1 answer:
Karolina [17]3 years ago
4 0

Answer:

The velocity is 40 ft/sec.

Explanation:

Given that,

Force = 3200 lb

Angle = 30°

Speed = 64 ft/s

The resistive force with magnitude proportional to the square of the speed,

F_{r}=kv^2

Where, k = 1 lb s²/ft²

We need to calculate the velocity

Using balance equation

F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}

Put the value into the formula

3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}

Put the value of k

3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}

1600-v^2=m\dfrac{d^2v}{dt^2}

At terminal velocity \dfrac{d^2v}{dt^2}=0

So, 1600-v^2=0

v=\sqrt{1600}

v=40\ ft/sec

Hence, The velocity is 40 ft/sec.

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