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+at.
 
Substituting the known values,
 
v=20m/s+(-2m/s2×4s)v=20-8v=12m/s
 
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.
 
Cheetah.jpg
 
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=ut+12at2.
 
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,
 
s=ut+12at250=0×t+(12×5×t2)250=12×5×t2250×2=5×t2500=5×t2t2=5005t2=100t=10seconds
 
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.
 
Metro.webp
 
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, v2=u2+2as.
 
Substituting the known values,
 
v2=u2+2asv2=0+2(3m/s2×150)mv2=2(3×150)v2=2(450)v2=900v=30m/s
 
Therefore, the train's velocity is 30 \(m/s\).