Physics Formula s=ut+1/2at^2

Physics Formula: \(s = ut + \frac{1}{2}at^2\) Explained with Examples

The equation \(s = ut + \frac{1}{2}at^2\) is one of the fundamental equations of motion in physics. It describes the displacement (\(s\)) of an object under uniform acceleration (\(a\)) over a period of time (\(t\)), given its initial velocity (\(u\)). This equation is derived from the kinematic principles of motion and is widely used in solving problems related to uniformly accelerated motion.

Understanding the Components of the Formula

The formula \(s = ut + \frac{1}{2}at^2\) can be broken down into the following components:

s: Displacement of the object (in meters).
u: Initial velocity of the object (in meters per second).
a: Uniform acceleration (in meters per second squared).
t: Time for which the object is in motion (in seconds).

Derivation of the Formula

The formula is derived from the basic definitions of motion under uniform acceleration:

The first equation of motion is: \[v = u + at\]
The average velocity (\(v_{avg}\)) is given by: \[ v_{avg} = \frac{u + v}{2} \] Substituting \(v = u + at\) into the above equation: \[ v_{avg} = \frac{u + (u + at)}{2} = u + \frac{1}{2}at \]
Displacement is the product of average velocity and time: \[ s = v_{avg} \cdot t \] Substituting \(v_{avg} = u + \frac{1}{2}at\): \[ s = \left(u + \frac{1}{2}at\right)t \] Simplifying: \[ s = ut + \frac{1}{2}at^2 \]

Applications of the Formula

The equation \(s = ut + \frac{1}{2}at^2\) is essential in solving problems involving:

Projectiles and free-falling objects.
Vehicle acceleration and deceleration.
Motion of objects on inclined planes.

Example Problems

Example 1: Free Fall

A ball is dropped from a height of 20 m with an initial velocity \(u = 0\). Calculate the time it takes to hit the ground. Assume \(a = 9.8 \, \text{m/s}^2\).

Solution:

Given: \[ u = 0, \, s = 20 \, \text{m}, \, a = 9.8 \, \text{m/s}^2 \]
Using the formula: \[ s = ut + \frac{1}{2}at^2 \] Substituting the values: \[ 20 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \] Simplifying: \[ 20 = 4.9t^2 \] \[ t^2 = \frac{20}{4.9} \approx 4.08 \] \[ t = \sqrt{4.08} \approx 2.02 \, \text{seconds} \]
Time to hit the ground: 2.02 seconds.

Example 2: Car Acceleration

A car starts from rest and accelerates uniformly at \(2 \, \text{m/s}^2\). Calculate the distance covered in 5 seconds.

Solution:

Given: \[ u = 0, \, a = 2 \, \text{m/s}^2, \, t = 5 \, \text{s} \]
Using the formula: \[ s = ut + \frac{1}{2}at^2 \] Substituting the values: \[ s = 0 \cdot 5 + \frac{1}{2} \cdot 2 \cdot 5^2 \] Simplifying: \[ s = 0 + \frac{1}{2} \cdot 2 \cdot 25 = 25 \, \text{meters} \]
Distance covered: 25 meters.

Example 3: Deceleration

A cyclist traveling at \(u = 10 \, \text{m/s}\) applies brakes, decelerating uniformly at \(a = -2 \, \text{m/s}^2\). Calculate the distance traveled before coming to rest.

Solution:

Given: \[ u = 10, \, v = 0, \, a = -2 \]
We know: \[ v^2 = u^2 + 2as \] Rearranging: \[ s = \frac{v^2 - u^2}{2a} \] Substituting: \[ s = \frac{0^2 - 10^2}{2 \cdot -2} = \frac{-100}{-4} = 25 \, \text{meters} \]
Distance before stopping: 25 meters.

Conclusion

The equation \(s = ut + \frac{1}{2}at^2\) is a vital tool in physics, helping us understand and predict the motion of objects under uniform acceleration. By applying this formula to real-world problems, you can analyze motion with precision and clarity.