## Little Paulie

### Task number: 2227

Little Paulie started off on a bike on a straight road. The acceleration dependence on time is given by the following table:

t / 10^{-1}s |
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

a / m·s^{-2} |
4,0 | 3,7 | 3,3 | 3,0 | 2,6 | 2,3 | 1,9 | 1,6 | 1,2 | 0,8 | 0,4 |

Draw a graph and determine the shape of the function *a*(*t*). Determine the course of Paulie’s speed and position dependence on time from it.

Suppose that Paulie was not moving at time *t* = 0 s and was at the origin of the coordinate system.

**Comment:** We should speak about the absolute value of acceleration and the absolute value of speed to be more precise. The movement is one-directional and its direction is not changing, thus we will speak about acceleration and speed.

#### Hint 1: Graph of Paulie’s acceleration dependence on time

Draw a graph of acceleration dependence on time from the given table.

What is the graph of the given dependence?

#### Hint 2: Expressing the function

*a*(*t*)Use the slope-intercept form of a line to express the function

*a*(*t*).#### Hint 3: Paulie’s speed time dependence

You know how Paulie’s speed changes with time. How to you find out his speed dependence on time?

#### Hint 4: Paulie’s position course

You already know the time dependence of Paulie’s speed.

How do you find his position dependence on time?

#### Solution

\[y\,=\,1\,\mathrm{m}-2\,\mathrm{m \cdot s^{-1}}\cdot t\,.\]**Comment:**Equations should be written for example as:We don’t write units for simplification.

We will draw a graph of Paulie’s acceleration dependence on time according to the table:

The graph is a line.

We will use the slope-intercept form of a line to express the function

\[a\left(t\right)\,=\,-kt+q\,,\]*a*(*t*):\(q\, =\, 4\) (intersection with the vertical axis)

\(k\,=\,tg\,\alpha\) (slope)

\[tg\alpha\,=\,\frac{\Delta{a}}{\Delta{t}} \,=\, \frac{4-0{,}4}{1} \,=\, 3{,}6\,.\](

*α*is the angle that the line clamps with the*x*-axis)The course of Paulie’s speed can be found by integration of

\[v\left(t\right)\,=\,\int{a\left(t\right)}dt\,=\,\int{\left(-3{,}6t+4\right)}dt\,,\] \[v\left(t\right)\,=\,-\frac{3{,}6}{2}t^{2}+4t+C\,=\,-1{,}8^{2}+4t+C\,.\]*a*(*t*):We can find the constant

*C*from the initial conditions:for

*t*= 0 s is*v*= 0 m s^{-1},thus

*C*= 0.We get:

\[v\left(t\right)\,=\,-1{,}8t^{2}+4t\,.\]The course of Pavlík’s speed can be found by integration of

\[s\left(t\right)\,=\,\int{v\left(t\right)}dt\,=\,\int{\left(-1{,}8t^{2}+4t\right)}dt\,=\, -\frac{1{,}8}{3}t^{3}+\frac{4}{2}t^{2}+D\,,\] \[s(t)\,=\, -0{,}6t^{3}+2t^{2}+D\,.\]*v*(*t*):We can find the constant

*D*from the starting conditions:for

*t*= 0 s is*s*= 0 m,thus

*D*= 0.#### Answer