Question

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 exerts a resistive force with magnitude proportional to the square of the speed with k

Answer 1

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.