# Distance, Velocity and Acceleration

## Solved Problems

Click or tap a problem to see the solution.

### Example 5

An object moves along a straight line with acceleration given by
\[a\left( t \right) = 1 + \cos \left( {\pi t} \right).\]
Assuming that the initial velocity at *t* = 0 is zero, find the total distance traveled over the first second.

### Example 6

A particle starts moving from rest with a constant acceleration. For the 1st second it covers *d*_{1} meters. What distance *d*_{2} does the particle cover for the 2nd second?

### Example 7

A particle starts from rest and moves with an acceleration given by the equation \[a\left( t \right) = {e^t} - 1\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right).\] Find the average speed of the particle over the interval from \({t_1} = 0\,\text{s}\) to \({t_2} = 2\,\text{s}.\)

### Example 8

A particle starts with the initial velocity of \({v_0} = 6\,\left( {\frac{\text{m}}{\text{s}}} \right)\) and acceleration \(5 - 2t\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right).\) Calculate the distance traveled by the particle for the first \(8\) seconds.

### Example 9

A particle starts from rest and moves with constant acceleration \({a_1} = 5\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right)\) for \({T_1} = 5\) seconds. It then decelerates uniformly for \({T_2} = 10\) seconds and stops. What is the distance covered by the particle?

### Example 10

A ball thrown vertically travels double the distance in the \(1\text{st}\) second than in the \(2\text{nd}\) second. What is the maximum height achieved by the ball?

### Example 5.

An object moves along a straight line with acceleration given by \[a\left( t \right) = 1 + \cos \left( {\pi t} \right).\] Assuming that the initial velocity at \(t = 0\) is zero, find the total distance traveled over the first second.

Solution.

First we compute the velocity of the object at time \(t.\) We use the formula

where \(u\) is a variable of integration. Hence,

Notice that the velocity \(v\left( t \right)\) is always positive for \(t \gt 0.\) Therefore, the total distance \(d\) traveled by the object for \(1\) sec is given by

The distance \(d\) is measured in meters, if the acceleration \(a\) is measured in meters per second squared.

### Example 6.

A particle starts moving from rest with a constant acceleration. For the \(1\text{st}\) second it covers \({d_1}\) meters. What distance \({d_2}\) does the particle cover for the \(2\text{nd}\) second?

Solution.

Let the acceleration of the particle be equal to \(a\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right)\). For motion from rest at constant acceleration, the velocity is given by the formula

The distance \({d_1}\) traveled by the particle for the \(1\text{st}\) second is written in the form

Similarly, we can write down the distance \({d_2}\) covered in the \(2\text{nd}\) second:

Comparing both expressions, we find that

where \({d_1}\) and \({d_2}\) are measured in meters.

### Example 7.

A particle starts from rest and moves with an acceleration given by the equation \[a\left( t \right) = {e^t} - 1\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right).\] Find the average speed of the particle over the interval from \({t_1} = 0\,\text{s}\) to \({t_2} = 2\,\text{s}.\)

Solution.

Integrating the acceleration function gives us the velocity function:

Since \({v_0} = 0,\) we have

Hence, the velocity function has the form:

The average speed of an object is defined as the total distance traveled by the object divided by total time. So, we can write

### Example 8.

A particle starts with the initial velocity of \({v_0} = 6\,\left( {\frac{\text{m}}{\text{s}}} \right)\) and acceleration \(5 - 2t\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right).\) Calculate the distance traveled by the particle for the first \(8\) seconds.

Solution.

We calculate the velocity function from the acceleration function:

To find \(C,\) we use the initial condition \({v_0} = 6\,\left( {\frac{\text{m}}{\text{s}}} \right),\) so we get

Notice that the velocity is equal to zero when \(t =6\) seconds. Hence, when calculating the distance, we split the interval of integration into two intervals where the velocity has a constant sign.

Thus, the total distance traveled by the particle is given by

### Example 9.

A particle starts from rest and moves with constant acceleration \({a_1} = 5\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right)\) for \({T_1} = 5\) seconds. It then decelerates uniformly for \({T_2} = 10\) seconds and stops. What is the distance covered by the particle?

Solution.

Determine the velocity \({v_1}\) at the moment \(t = {T_1}:\)

On the second stage, the particle's velocity varies according to the equation

where \(0 \le t \le {T_2}.\) Since \(v\left( {{T_2}} \right) = 0,\) we can easily find the value of acceleration \({a_2}:\)

The value of \({a_2}\) is negative as the particle is decelerating.

Let's now move on to the calculation of distances. On the first stage, the distance traveled by the particle is equal to

The distance traveled on the second stage is given by

Hence, the total distance covered by the particle is

### Example 10.

A ball thrown vertically travels double the distance in the \(1\text{st}\) second than in the \(2\text{nd}\) second. What is the maximum height achieved by the ball?

Solution.

Suppose that the ball is thrown upward with an initial velocity of \({v_0}.\)

The velocity of the ball varies by the law

where \(g\) is the acceleration due to gravity: \(g \approx 9.8\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right)\)

The distance traveled in the \(1\text{st}\) second is

Respectively, the distance covered for the \(2\text{nd}\) second is given by

We can find \({v_0}\) from the condition \(\frac{{{d_1}}}{{{d_2}}} = 2.\) So

The maximum height is determined by the equation

Hence,