Answer
$\left|v\right|=15.8m/s$
Work Step by Step
$a_{x}=5.00m/s^{2}$ is constant, so we can use Table 2-1 in the x-direction.
$a_{y}=7.00m/s^{2}$ is constant, so we can use Table 2-1 in the y-direction..
$\left[\begin{array}{ll}
v=v_{0}+at, & (2- 11)\\
x-x_{0}=v_{0}t+\frac{1}{2}at^{2}, & (2- 15) \\
v^{2}=v_{0}^{2}+2a(x-x_{0}) , & (2- 16) \\
x-x_{0}=\frac{1}{2}(v_{0}+v)t, & (2- 17) \\
x-x_{0}=vt-\frac{1}{2}at^{2} & (2- 18)
\end{array}\right]$
Given $x-x_{0}=12.0m, \quad v_{x0}=4.0m/s,\quad a_{x}=5.00m/s^{2},$
we solve $(2- 15)$ for $t:$
$12.0 m=(4.00m/s)t+\displaystyle \frac{1}{2}(5.00m/\mathrm{s}^{2})t^{2}$
$(2.50m/\mathrm{s}^{2})t^{2}+(4.00m/s)t-12.0 m=0$
The quadratic formula gives
$t=\displaystyle \frac{(4.00m/s)\pm\sqrt{(4.00m/s)-4(-12.0 m)(2.50m/\mathrm{s}^{2})}}{2(2.50m/\mathrm{s}^{2})}$
discarding the negative solution, gives $t=1.53\mathrm{s}$.
We now find components of velocity at $t=1.53\mathrm{s}$.
In the x direction , Eq. 2-11 gives
$v_{x}=v_{x0}+a_{x}t=4.00m/s+(5.00m/s^{2})(1.53\mathrm{s})=11.7m/s$
In the $y$ direction , Eq. 2-11 gives
$v_{y}=v_{y0}+a_{y}t=0+(7.00m/s^{2})(1.53\mathrm{s})=10.7m/s$
Thus, $\qquad \vec{v}=(11.7m/s)\hat{i}+(10.7m/s)\hat{j}$
(a)
Its magnitude is
$\left|v\right|=\sqrt{(11.7m/s)+(10.7m/s)}=15.8m/s$
