2D Kinematics Problem-solving

Kinematics is used in astrophysics to describe the motion of celestial bodies and collections of such bodies. In mechanical engineering, robotics, and biomechanics, kinematics is used to describe the motion of systems composed of joined parts (multi-link systems) such as an engine, a robotic arm, or the human skeleton.

The most important thing to remember in 2D kinematics problems is that the two dimensions are entirely independent of each other. So, that means you are never actually doing a 2D kinematics problem, but you are always doing two 1D kinematics problems at the same time. Literally, it is the same in the two problems; that's all that connects them.

An airplane is taking off on the runway. At the moment the wheels leave the ground, the plane is traveling at $60\text{ m/s}$horizontally. The wings generate a lift, which causes a vertical acceleration of $1\text{ m/s}^2,$and the jets produce a horizontal acceleration of $2\text{ m/s}^2.$When the airplane is flying at an altitude of $3200\text{ m},$what is its horizontal distance from the airport?

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Suppose the plane starts from rest and gains altitude (ignore the horizontal traveling of the plane for now). Then using the formula

$v^2-u^2 = 2as,$

we can get the vertical velocity.

As the plane starts from rest, the initial velocity $u$will be zero for the plane. Therefore,

$v^2 =2as.$

Since $a \text{ (vertical)} =1 \text{ m/s}^2$and $s = 3200\text{ m},$it follows that $v= 80\text{ m/s}.$

Use this $v$to obtain the time of flight:

$\begin{aligned} v&=u +at \\ &=0+at\\\\ \Rightarrow t&= 80 \text{ sec}. \end{aligned}$

Now, by the formula

$s = u \text{ (horizontal)}\, t + \frac{1}{2} at^2,$

the horizontal distance is

$s = 60\times 80 + \frac{1}{2}\times 2\times 80^2 = 11200\text{ m}.\ _\square$