Equations of motion

Four equations of motion are used to describe the motion of an object that is moving with a constant acceleration.

These equations relate an object’s displacement, velocity, acceleration, and time.
Here is an explanation of each equation:

v = u + at
where
u – initial velocity
v – final velocity
a – acceleration
t – time.
It shows that an object’s final velocity equals its initial velocity plus the product of its acceleration and the time elapsed.

s = ut + 1/2 at²
where
s – object’s displacement
u – initial velocity
a – acceleration
t – time
It shows that an object’s displacement equals the product of its initial velocity and time plus half the product of its acceleration and the square of the time elapsed.

v² = u² + 2as
where
u -initial velocity
v – final velocity
a – acceleration
s – displacement
It shows that an object’s final velocity squared equals its initial velocity squared plus twice the product of its acceleration and displacement.

s = (u+v) t /2
where
u -initial velocity
v – final velocity
a – acceleration
s – displacement
It shows that an object’s displacement equals the sum of its initial and final velocity multiplied by time, divided by 2

Example:
Calculate the final velocity of a car that starts from rest and accelerates at a constant rate of 2 m/s² for 5 seconds.

s = ?
u = 0 m/s
v = ? m/s
a = 2 m/s²
t – 5 s

You can use the second equation to calculate the distance traveled by the same car during that 5-second interval.
s = ut + 1/2 at² = 0 x 5 + 1/2 x 2 x 5² = 25 m
Finally, you can use the third equation to calculate the final velocity of the car if it travels a distance of 50 meters and accelerates at a constant rate of 5 m/s².

v² = u² + 2as = 0² + 2 x 2 x 25 = 100
v = 10 m/s

These equations are only valid for motion with constant acceleration, however in real-life scenarios often involve more complex forms of motion.
Nevertheless, the three equations of motion are important tools for understanding and predicting the behavior of objects in motion under certain conditions.

More examples:

A car is moving with an initial velocity of 20 m/s. After 10 seconds, its velocity becomes 40 m/s. What is the car’s acceleration?
A ball is thrown vertically upward with an initial velocity of 30 m/s. How high will it go before falling back to the ground? (Assume negligible air resistance)
A car accelerates from rest and reaches a speed of 50 m/s in 10 seconds. How far has it traveled during this time?
A stone is dropped from the top of a building that is 100 meters tall. How long will it take for the stone to hit the ground? (Assume negligible air resistance)
A train accelerates from rest at a rate of 5 m/s^2. How long will it take for the train to reach a speed of 100 m/s?

Solutions:

Acceleration (a) = (final velocity – initial velocity) / time

2. The time taken for the ball to reach its maximum height (h) is given by the equation
v = u + at
where
u = 30 m/s
a = -9.8 m/s² (due to gravity)
v = 0 m/s (at maximum height).
Solving for t, we get

The maximum height reached by the ball is given by the equation
h = ut + 1/2 at²,
where
u = 30 m/s
a = -9.8 m/s²
t = 3.06 s
Plugging in these values
h = ut + 1/2 at² = 30 x 3.06 + 1/2 x (-9.8) x 3.06² ≈ 46.5 m.

3. The displacement (s) of the car is given by the equation
s = ut + 1/2 at²
where
u = 0 m/s (since the car starts from rest),
a = 5 m/s²,
t = 10 s.
Plugging in these values, we get
s = 0 m + 1/2 x 5 x 10² = 250 m.

4. The time taken for the stone to hit the ground (t) is given by the equation
s = ut + 1/2 at²
s= h = 0 x t + 1/2 x 9.8 x t²,
where
h = 100 m, g = 9.8 m/s², and t is the unknown.
Plugging in these values, we get
100 = 1/2 x 9.8 x t²
t² = 100 / 4.9
t≈ 4.52 s

5. The time taken for the train to reach a speed of 100 m/s is given by the equation
v = u + at,
where
u = 0 m/s (since the train starts from rest),
a = 5 m/s²,
v = 100 m/s.
Solving for t, we get
t = (v-u)/a = (100 -0) / 5 = 20 s.