## Total Distance Traveled Formula

Distance traveled describes how much path an object has covered in order to reach its destination in a specified time period. The distance covered formula for distance traveled is given as:

$d = vt$

Where,

$d$ = the distance traveled

$v$ = the velocity

$t$ = time taken to travel

Similar to the distance traveled formula, there is the distance traveled in the last second formula and formula for displacement which we will learn below.

## Uses and Applications of Distance Traveled Formula

The distance-covered formula is applicable to compute the distance of driving a car or swimming stretch in a pool. While driving a car, the distance will be computed in kilometers or miles, the rate is in kilometers per hour or miles per hour, and time is in hours. While swimming laps in a pool, the distance is computed in laps.

## Displacement in nth Second Formula

In order to calculate the displacement (position shift) from the velocity function, you just need to integrate the function. The negative areas below the x-axis subtracted from the total displacement. For this, we use a formula for displacement in nth second.

Displacement $= \int b_a v(t) dt$

In order to calculate the distance traveled we need to use the absolute value.

The displacement in the nth second distance formula covered in the nth second is given by $S_n = u + a(n - 12)$.

## Derive the Expression for the Formula for Displacement in the nth Second

A is the acceleration

V is the velocity

U is the initial velocity

S is the distance

$s = ut + \dfrac{1}{2}at^2$

In order to calculate the distance traveled at the time of the nth second, we compute the distance covered in n seconds and subtract (minus-) the distance covered in (n-1) seconds and obtain:-

$s = un - u(n-1) + \dfrac{1}{2}an^2 - \dfrac{1}{2}a(n-1)^2$

Simplifying provides us

$u \text{ term }= u(n-(n-1)) = u \times 1 = u$

$(n-1)^2 = n^2-2n+1$

$a \text{ term }= \dfrac{1}{2}a(n^2-(n-1)^2) = \dfrac{1}{2}a(n^2-n^2+2n-1)$ the $n^2$'s cancel and provide us $\dfrac[1}{2a}(2n-1)$.

The final distance traveled equation for displacement in the nth second is

s = u + \dfra{1}{2a (2n-1)}$

## Solved Examples

Example: A heavily loaded bus travels at a velocity of 80 miles per hour. Find out the total time taken by the truck to travel a distance of 300 miles?

Solution

Known: Velocity = 80 miles per hour

Displacement d = 300 miles

We know that,

Displacement $d = vt$

The time taken is provided by:-

$t = \dfrac{d}{v}$

$t = \dfrac{300}{80}$

$t = 3.75$ hours

## Conclusion:

The distance formula is a method for calculating the distance between two places. These points can be of any size and in any dimension. Every day, people deal with the distance between two points. Inches, feet, miles, millimeters, and meters can all be used to describe it. The distance covered formula for distance traveled is given as: $d = vt$

## FAQs on Distance Traveled Formula

**1. How to Calculate Time and Distance from Velocity and Acceleration?**

Given a constant acceleration and the alteration in velocity of an object, you can evaluate both the distance traveled and the time involved. For example, imagine you’re a drag racer. Your acceleration will be 24.6 meters per second square, and your ultimate speed is 164.3 meters per second. Now find the total distance traveled.

You are aware of the acceleration and the ultimate speed, and you seek to know the total distance needed to get to that speed. You are aware of the final speed, $v_f$. and the initial speed, $v_i$ (which is 0), and you know the acceleration, a. Since $v_t - v_i = at$, you know that:-

$T = v_f – \dfrac{v_i}{a}$

$T = \dfrac{(164.3 - 0)m/s}{24.6 m/s}$

$T = 6.67s$

**2. What is the distance traveled in the last second formula?**

Thinking about how we find displacement in the last two seconds?

Initial velocity of body $u = 0$.

Using $v = u + at$.

We obtain $v = 0 + an$.

For displacement in the last two seconds, $t=2$.

Taking into account $S= vt − 2at^2$.

Or $S = v (2) −2nv (2)^2$

$S = 2v - 8nv$

$S = 2v(1-2^2n)$