#### Known

For an object moving with constant acceleration, we have the following equations of motion.

$$
begin{align}
&v_f=v_i+acdot t\
&x_f=x_i+v_icdot t+frac{1}{2}cdot acdot t^2
end{align}
$$

Where $x_i$ is the initial position of the object and $v_i$ is the velocity it has at the moment it begins to accelerate, that is, the initial velocity.

The scooter is stopped at the traffic light and then begins to accelerate. Therefore its initial position and its initial velocity are $x_i=0 tx{m}$ and $v_i=0 frac{tx{m}}{tx{s}}$ respectively.
#### Calculation

Givens: $a=4.2 frac{tx{m}}{tx{s}^2}$, $t=3.0 tx{s}$, $x_i=0 tx{m}$, $v_i=0 frac{tx{m}}{tx{s}}$.

a) From (1) we have:

$$
begin{align*}
v_f=0 frac{tx{m}}{tx{s}}+left(4.2 frac{tx{m}}{tx{s}^2}right)(3.0 tx{s})=12.6 frac{tx{m}}{tx{s}}
end{align*}
$$

$$
begin{align*}
boxed{v_f=12.6 frac{tx{m}}{tx{s}}}
end{align*}
$$

b) From (2) we have:

$$
begin{align*}
x_f=0 tx{m}+left(0 frac{tx{m}}{tx{s}}right)(3.0 tx{s})+frac{1}{2}left(4.2 frac{tx{m}}{tx{s}^2}right)(3.0 tx{s})^2=18.9 tx{m}
end{align*}
$$

$$
begin{align*}
boxed{x_f=18.9 tx{m}}
end{align*}
$$

—

#### Conclusion

After 3 $tx{s}$ the scooter is 18.9 $tx{m}$ from where it was and is moving with a speed of $12.6 frac{tx{m}}{tx{s}}$.

subsection*{Known}

The soccer ball begins to accelerate due to gravity, that is, acceleration and velocity have the same direction. Considering constant acceleration, we have the following equation for position, average velocity, and time.
begin{align}
&x_f=x_i+left(frac{v_i+v_f}{2}right)tnotag\
&implies boxed{v_f=frac{2(x_f-x_i)}{t}-v_i}
end{align}
subsection*{Calculation}
Givens: $v_i=0 frac{tx{m}}{tx{s}}$, $x_i=0 tx{m}$, $x_f=6.5 tx{m}$, $t=3.1 tx{s}$.
Using (1):
begin{align*}
v_f=frac{2(6.5 tx{m}-0 tx{m})}{3.1 tx{s}}-0 frac{tx{m}}{tx{s}}=4.2 frac{tx{m}}{tx{s}}
end{align*}
begin{align*}
boxed{v_f=4.2 frac{tx{m}}{tx{s}}}
end{align*}

vspace{2pt}
hrule
vspace{1pt}

subsection*{Conclusion}

The speed of the soccer ball is $4.2 frac{tx{m}}{tx{s}}$.\

Note that speed is a scalar.

subsection*{Known}

The bicycle’s brakes cause the bicycle to begin to decelerate, that is, acceleration and velocity have opposite directions. Considering constant acceleration, we have the following equation for position, average velocity and time.

begin{align}
&x_f=x_i+left(frac{v_i+v_f}{2}right)tnotag\
&implies boxed{t=frac{2(x_f-x_i)}{v_i+v_f}}
end{align}
subsection*{Calculation}
Givens: $v_i=6.4 frac{tx{m}}{tx{s}}$, $v_f=0 frac{tx{m}}{tx{s}}$ ,$x_i=0 tx{m}$, $x_f=3.8 tx{m}$.\

Using (1)
begin{align*}
t=frac{2(3.8 tx{m}-0 tx{m})}{6.4 frac{tx{m}}{tx{s}}+0 frac{tx{m}}{tx{s}}}=1.2 tx{s}
end{align*}

begin{align*}
boxed{t=1.2 tx{s}}
end{align*}

vspace{2pt}
hrule
vspace{1pt}

subsection*{Conclusion}

It would take 1.2 s to stop.