Answer
$(a)\space 25.5\space m/s$
$(b)\space 34.4^{\circ}$ from $+y\rightarrow +x$ direction
Work Step by Step
(a) The vehicle accelerates only in the +x direction. Therefore the y component of its velocity remains constant at 21 m/s.
$V_{y}=21\space m/s$
Let's apply equation 3.3a $V=u+at$ in the horizontal direction to find the x-component of the velocity after 45s.
$\rightarrow V=u+at$ ; Let's plug known values into this equation.
$V_{x}=0+0.320\space m/s\times 45\space s=14.4\space m/s$
According to the Pythagorean theorem, we can get the magnitude of the vehicle's velocity after 45s.
$V=\sqrt {V_{x}^{2}+V_{y}^{2}}=\sqrt {(14.4\space m/s)^{2}+(21\space m/s)^{2}}=25.5\space m/s$
(b) By using trigonometry, we can get the angle of the vehicle's velocity with respect to the +y direction.
$tan\theta=\frac{V_{x}}{V_{y}}=\frac{14.4\space m/s}{21\space m/s}$
$\theta=tan^{-1}(0.69)=34.4^{\circ}$
