### Theory:

Let us see few examples to understand the application of the three equations of motion.

1. Practical problem using the first equation of motion:

A car cruise down an expressway at 20 $$m/s$$. Engineers are designing an off-ramp in an interchange with a deceleration of −2 $$m/s²$$ that lasts 4 $$s$$. Then, what would be the velocity of cars at the end of the off-ramp?

Important!
A ramp, also known as an inclined plane, is a flat supporting surface tilted at an angle with one end angle is higher. It is mostly used as an aid for raising or lowering a road.
Given:

Initial velocity of car, $$u =$$ 20 $$m/s$$

Acceleration of the car, $$a =$$ −2 $$m/s²$$

Time, $$t =$$ 4 $$s$$

To find the final velocity of the car, we can use the first equation of motion, $v=u+\mathit{at}$.

Substituting the known values,

$\begin{array}{l}v=20\phantom{\rule{0.147em}{0ex}}m/s+\left(-2\phantom{\rule{0.147em}{0ex}}m/{s}^{2}×4\phantom{\rule{0.147em}{0ex}}s\right)\\ \\ v=20\phantom{\rule{0.147em}{0ex}}-8\\ \\ v=12 m/s\end{array}$

Therefore, the car has a velocity of  12 $$m/s$$ at the end of the off-ramp.

2. Practical problem using the second equation of motion:

Consider a cheetah starts from rest and accelerate uniformly at 5 $$m/s²$$ for 250 $$m$$. Find out the time it takes to cover this distance. Given:

Since it starts from the rest, the initial velocity of cheetah, $$u = 0$$ $$m/s$$

Acceleration of the cheetah, $$a =$$ 5 $$m/s²$$

Displacement, $$s =$$ 250 $$m$$

To find the time to cover this distance, we can use the second equation of motion,  $s=\mathit{ut}+\frac{1}{2}a{t}^{2}$.

Since the displacement is given and we do not have the velocity of the cheetah, we cannot use the first equation of motion.

Substituting the known values,

$\begin{array}{l}s=\mathit{ut}+\frac{1}{2}\mathit{at}\\ \\ 250=0×t+\left(\frac{1}{2}×5×{t}^{2}\right)\\ \\ 250=\frac{1}{2}×5×{t}^{2}\\ \\ 250×2=5×{t}^{2}\\ \\ 500=5×{t}^{2}\\ \\ {t}^{2}\phantom{\rule{0.147em}{0ex}}=\frac{500}{5}\\ \\ {t}^{2}=100\\ \\ t\phantom{\rule{0.147em}{0ex}}=10\phantom{\rule{0.147em}{0ex}}\mathit{seconds}\end{array}$

Therefore, cheetah will cover 250 $$m$$ in 10 $$s$$ if it accelerates uniformly at 5 $$m/s²$$.

3. Practical problem using the third equation of motion:

A 10 car metro train is at rest in a station. It reaches its cruising speed after accelerating at 3 $$m/s²$$ for a distance equivalent to the length of the station 150 $$m$$. It then travels at a constant speed towards the next station 20 blocks away for ($$1325\ m$$). In such a case, determine the train's cruising velocity. Given:

The train starts from the rest, so the initial velocity of train $$u = 0$$ $$m/s$$.

Acceleration of the train, $$a =$$ 3 $$m/s²$$

Displacement of the train (length of the station), $$s =$$ 250 $$m$$

To find the velocity of the train, we can use the third equation of motion, ${v}^{2}={u}^{2}+2\mathit{as}$.

Substituting the known values,

$\begin{array}{l}{v}^{2}={u}^{2}+2\mathit{as}\\ \\ {v}^{2}=0\phantom{\rule{0.147em}{0ex}}+2\phantom{\rule{0.147em}{0ex}}\left(3\phantom{\rule{0.147em}{0ex}}m/{s}^{2}×150\right)\phantom{\rule{0.147em}{0ex}}m\\ \\ {v}^{2}=2\phantom{\rule{0.147em}{0ex}}\left(3×150\right)\\ \\ {v}^{2}=2\phantom{\rule{0.147em}{0ex}}\left(450\right)\\ \\ {v}^{2}=900\\ \\ v\phantom{\rule{0.147em}{0ex}}=30\phantom{\rule{0.147em}{0ex}}m/s\end{array}$

Therefore, the train's velocity is 30 $$m/s$$.