Answer
Please see the work below.
Work Step by Step
As we know that
$\Delta v=v_2-v_1$
$\Delta v=\sqrt{v_1^2+v_2^2-2v_1v_2cos(\theta)}$
We plug in the known values to obtain:
$\Delta v=\sqrt{(15)^2+(19)^2-2(15)(19)cos(28)}$
$\Delta v=9.095\frac{Km}{s}$
$\Delta v=9095\frac{m}{s}$
Now we can find the acceleration as
$a=\frac{\Delta v}{\Delta t}$
$a=\frac{9095}{10\times 60}=15\frac{m}{s^2}$
