In Physics **Make a Subject Of The Formula v = u + at: **this formula is commonly used in mechanics and is one of the fundamental equations that describe the motion of objects. It can be applied to a wide range of situations, from the motion of cars on the highway to the movement of planets in space.

In this formula, the acceleration can be positive or negative, depending on the direction of the force acting on the object. For example, if a car is accelerating forward, the acceleration would be positive, whereas if it is slowing down, the acceleration would be negative.

By rearranging the formula, it is also possible to calculate other variables such as the initial velocity (u), the acceleration (a), or the time taken (t), depending on what information is available.\

**In Mathematics **

To make “a” the subject of the formula v = u + at, we need to isolate the variable “a” on one side of the equation.

Starting with the original formula:

v = u + at

Subtracting “u” from both sides:

v – u = at

Dividing both sides by “t”:

a = (v – u) / t

Therefore, the value of acceleration “a” can be determined by taking the difference between the final velocity “v” and the initial velocity “u”, and then dividing the result by the time “t”.

Further: This article will cover the process of rearranging algebraic equations to solve for specific variables. Specifically, I will demonstrate how to solve for u in equation **v = u + at, solve for t in the equation m = t + 1/t-3, and solve for u in the equation v2 = u2 + 2as.**

These types of questions require some basic algebraic skills such as addition, subtraction, cross multiplication, squaring, taking square roots, collecting like terms, and rearranging variables. I will provide clear and easy-to-follow steps for each equation and am available to answer any questions or provide further clarification.

**Also Solved: Arithmetic Progression â€“ Nth Term, AP Formula, Sum of AP & Example**

## Make a Subject Of The Formula v = u + at

Starting with the formula:

v = u + at

where v is final velocity, u is initial velocity, a is acceleration, and t is time.

To make a subject of the formula means to isolate one of the variables (in this case, u) on one side of the equation, with all the other variables on the other side.

To do this, we can start by isolating the term with the variable we want to make the subject of the formula (in this case, u) on one side of the equation:

v – at = u

This is the solution for u, and we can now substitute in values for v, a, and t to solve for u.

For example, if we have v = 20 m/s, a = 5 m/s^2, and t = 3 seconds, we can substitute those values into the formula to find u:

u = 20 m/s – (5 m/s^2)(3 s)

u = 20 m/s – 15 m/s

u = 5 m/s

So the initial velocity u is 5 m/s, given the final velocity of 20 m/s, acceleration of 5 m/s^2, and time of 3 seconds.

**Make u the subject of the formula v2 = u2 + 2as**

To make u the subject of the formula v^2 = u^2 + 2as, we need to isolate u on one side of the equation with all the other variables on the other side.

First, we can start by moving the 2as term to the other side of the equation by subtracting 2as from both sides:

v^2 – 2as = u^2

Next, we can take the square root of both sides of the equation to isolate u:

u = Â±âˆš(v^2 – 2as)

Note that there are two possible values for u because of the Â± sign. This is because taking the square root of a number can give a positive or negative result.

To determine which value of u to use, we need to consider the physical situation. If the object is moving in the same direction as the positive axis, we take the positive square root. If it is moving in the opposite direction, we take the negative square root.

For example, let’s say an object is dropped from rest and falls freely under gravity for 2 seconds, with an acceleration of -9.8 m/s^2. We can use the formula to find the final velocity of the object, and then use that to find its initial velocity:

Given: a = -9.8 m/s^2 (acceleration due to gravity) s = 0 (initial position is the same as final position, as the object is dropped from rest) t = 2 s (time the object falls) v = ? (final velocity)

Using the formula: v^2 = u^2 + 2as

We can rearrange it to solve for v: v^2 = 2as v = âˆš(2as)

v = âˆš(2 * -9.8 m/s^2 * 0) v = 0 m/s

So the final velocity of the object is 0 m/s.

Now we can use the formula for u to find the initial velocity: u = Â±âˆš(v^2 – 2as)

Using the positive square root since the object is falling in the positive direction: u = âˆš(0 m/s – 2(-9.8 m/s^2)(0))

u = âˆš0

u = 0 m/s

So the initial velocity of the object is also 0 m/s.

**Make t the subject of the equation m =t + 1/t-3**

To make t the subject of the equation, we need to isolate t on one side of the equation.

Starting with the given equation:

m = t + 1/(t – 3)

Multiply both sides of the equation by (t – 3):

m(t – 3) = t(t – 3) + 1

Expand the brackets on the right-hand side:

m(t – 3) = t^2 – 3t + 1

Rearrange the terms:

t^2 – 3t – m(t – 3) + 1 = 0

We can solve for t using the quadratic formula:

t = [3 Â± sqrt(9 + 4m(t – 3))]/2

Simplifying this expression:

t = (3 + sqrt(9 + 4m(t – 3)))/2 or t = (3 – sqrt(9 + 4m(t – 3)))/2

However, we can see that this expression involves t on both sides, which means it cannot be solved directly. Therefore, we need to use an iterative method to approximate the solution.

One such method is the Newton-Raphson method, which involves repeatedly applying the formula:

t_n+1 = t_n – f(t_n)/f'(t_n)

where t_n is the current approximation of t, f(t) is the function we want to solve (in this case, f(t) = m – t – 1/(t – 3)), and f'(t) is its derivative.

We can start with an initial guess for t, and then iteratively apply the formula until we reach a desired level of accuracy.

For example, let’s start with t_0 = 0 (this is just an arbitrary choice):

f(t_0) = m – t_0 – 1/(t_0 – 3) = m + 1/3

f'(t_0) = 1 + 1/(t_0 – 3)^2 = 10

Using these values in the formula, we get:

t_1 = t_0 – f(t_0)/f'(t_0) = 0 – (m + 1/3)/10

We can continue this process, using t_1 as the new approximation:

f(t_1) = m – t_1 – 1/(t_1 – 3)

f'(t_1) = 1 + 1/(t_1 – 3)^2

t_2 = t_1 – f(t_1)/f'(t_1)

and so on, until we reach the desired level of accuracy.

Note that this process may not always converge, depending on the value of m and the initial guess for t. In some cases, it may be necessary to use a different initial guess or a different method altogether.