Given,
The acceleration of the car,
[tex]a=2.5m/s^2[/tex]
The angle of inclination,
[tex]\theta=5.9^o[/tex]
Let the distance travelled along inclined be s at time t=15 s
Thus,
[tex]\begin{gathered} s=\frac{1}{2}at^2 \\ \Rightarrow s=\frac{1}{2}\times2.5\times15^2=281.25\text{ m} \end{gathered}[/tex]
Along horizontal the distance travelled is
x=scos5.9=281.25cos5.9=279.76 m
