Given data:
* The acceleration of the car is,
[tex]a=2ms^{-2}^{}[/tex]* The angle of the inclined plane is 5.5 degree.
* The time taken by the car is 12 s.
* The initial velocity of the car is 0 m/s.
Solution:
By the kinematics equation, the final velocity of the car on the inclined plane is,
[tex]v-u=at[/tex]where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the tiem taken,
Subsituting the known values,
[tex]\begin{gathered} v-0=2\times12 \\ v=24\text{ m/s} \end{gathered}[/tex]By the kinematics equation, the distance tarveled by the car on the inclined plane is,
[tex]v^2-u^2=2aS[/tex]where S is the distance tarveled on the inclined plane,
Substituting the known values,
[tex]\begin{gathered} 24^2-0=2\times2\times S \\ 576=4\times S \\ S=\frac{576}{4} \\ S=144\text{ m} \end{gathered}[/tex]The diagrammatic representation of the car on the inclined plane is,
The distance traveled by the car in the horizontal direction or along x-axis is,
[tex]\begin{gathered} S_x=S\cos (5.5^{\circ}) \\ S_x=144\times\cos (5.5^{\circ}) \\ S_x=143.34\text{ m} \end{gathered}[/tex]Thus, the distance tarveled by the car in the horizontal direction is 143.34 meter.
The distance tarveled by the car in the vertical direction or y-axis is,
[tex]\begin{gathered} S_y=S\sin (5.5^{\circ}) \\ S_y=144\times\sin (5.5^{\circ}) \\ S_y=13.8\text{ m} \end{gathered}[/tex]Thus, the distance tarveled by the car in the vertical direction is 13.8 meter.
