# Distance, Velocity and Acceleration

## Basic Relationships

Recall from our study of derivatives that if *x* (*t*) is the position of an object moving along a straight line at time *t*, then the velocity of the object is

and the acceleration is given by

Using the integral calculus, we can calculate the velocity function from the acceleration function, and the position function from the velocity function.

The Fundamental Theorem of Calculus says that

Similarly, the difference between the position at time \({{t_1}}\) and the position at time \({{t_2}}\) is determined by the equation

If the object moves from the position \(x\left( {{t_1}} \right)\) to the position \(x\left( {{t_2}} \right),\) the change \(x\left( {{t_2}} \right) - x\left( {{t_1}} \right)\) is called the displacement of the object:

To find the total distance traveled by the object between time \({{t_1}}\) and time \({{t_2}},\) we need to compute the integral of \(\left| {v\left( t \right)} \right|:\)

## Constant Acceleration

Suppose that an object is moving along a straight line with the constant acceleration \(a.\) At time \({t_1} = 0,\) the object has an initial velocity \({v_0}\) and an initial position \({x_0}.\)

The velocity and position of the object at time \(t\) are given by the equations

## Solved Problems

### Example 1.

The velocity of an object is given by the equation \[v\left( t \right) = \sqrt {4 + t} ,\] where the velocity \(v\) is measured in \(\frac{\text{m}}{\text{s}},\) the time \(t\) is measured in seconds. Find the distance traveled by the object for the first \(5\,\text{sec}.\)

Solution.

Given that the velocity is positive for \(t \gt 0,\) the total distance traveled for the time interval \(\left[ {0,t} \right)\) is expressed by the integral:

where \(u\) is the inner variable which has no impact on the computation of the integral.

Integration yields:

Substituting \(t = 5,\) we have

### Example 2.

A particle moves along a straight line in the positive direction of the \(x-\)axis with velocity given by the equation \[v = 2\sqrt{x}\,\left({\frac{\text{m}}{\text{s}}}\right).\] Assuming that \(x\left( {t = 0} \right) = 0,\) find the time \(t\) the particle takes to cover the first \(100\,\text{m}\) of the path.

Solution.

The equation of motion of the particle has the form

We have a simple differential equation that describes the particle's position as a function of time. Separating the variables and integrating both sides yields:

It follows from the initial condition \(x\left( {t = 0} \right) = 0\) that the \(C = 0.\) Hence, the particle moves according to the law:

It is easy to see that \(t = 10\,\text{s}\) when \(x = 100\,\text{m}.\)

### Example 3.

Starting at time \(t = 0,\) an object moves along a straight line with the velocity \[v\left( t \right) = 6 - 2t.\] Calculate the displacement and distance traveled by the object at time \(t = 5\,\text{s}.\)

Solution.

Let's first compute the displacement \(\Delta x\) of the object when \(t = 5\,\text{s}.\) Integrating the velocity expression, we obtain

Notice that the velocity changes sign at \({t_1} = 3.\) Therefore, to calculate the distance \(d\) traveled by the object we split the initial interval \(\left[ {0,5} \right]\) into two subintervals \(\left[ {0,3} \right]\) and \(\left[ {3,5} \right].\) This yields

### Example 4.

A particle starts from rest with an acceleration \(a\left( t \right)\) which varies according to the equation \[a\left( t \right) = \cos \frac{{\pi t}}{6}\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right).\] Find the distance traveled by the particle for the \(3\text{rd}\) second.

Solution.

Given that the initial velocity is zero: \({v_0} = 0,\) we determine the velocity equation:

The distance traveled in the \(3\text{rd}\) second is