91Ó°ÊÓ

Escape roads use gravitational acceleration to stop the car in an emergency.

On these roads, the component of gravitational acceleration acts against the motion to reduce the speed of the car.

Given data:

The initial speed of the truck is

$\begin{array}{c}{v}_{\mathrm{i}}=110\frac{\mathrm{km}}{\mathrm{h}}\\ =30.56\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

The final speed of the car is $v_{\mathrm{f}}=0$as it comes to rest.

The truck comes to rest within the time $t=7.0\mathrm{s}$.

The angle of the slope is $\mathrm{θ}=26°$.

Assumptions:

Let a be the acceleration of the truck.

Now, you know that

$\begin{array}{c}a=\frac{v_{\mathrm{f}}-v_{\mathrm{i}}}{t}\\ =\frac{0-30.56\frac{\mathrm{m}}{\mathrm{s}}}{7.0\mathrm{s}}\\ =-4.37\frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

The truck comes to rest. Therefore, the acceleration is against the direction of motion. Then, the direction of the acceleration is down the slope.

The horizontal component of the acceleration is

$\begin{array}{c}{a}_{\mathrm{h}}=a\cos \mathrm{θ}\\ =\left(-4.37\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\cos 26°\\ =-3.93\frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

The vertical component of the acceleration is

$\begin{array}{c}{a}_{\mathrm{v}}=a\sin \mathrm{θ}\\ =\left(-4.37\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\sin 26°\\ =-1.92\frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

Hence, the horizontal component of the acceleration is $-3.93\frac{\mathrm{m}}{{\mathrm{s}}^2}$, and the vertical component of the acceleration is $-1.92\frac{\mathrm{m}}{{\mathrm{s}}^2}$.