# In anticipation of a long 7° upgrade

In anticipation of a long 7° upgrade, a bus driver accelerates at a constant rate of 3 ft/$s^2$ while still on a level section of the highway. Knowing that the speed of the bus is 60 mi/h as it begins to climb the grade and that the driver does not change the setting of his throttle or shift gears, determine the distance traveled by the bus up the grade when its speed has decreased to 50 mi/h.

#### Solution:

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Let us first draw a free body diagram at the ground level depicting the forces affecting the bus.

Using Newton’s second law, we can write the following.

$+\uparrow\sum F=ma$

substitute $m=\dfrac{W}{g}$ and $F=p$ in our diagram

$p=\dfrac{W}{g}a$

Now, let us draw a free body diagram with the bus on the hill.

Again, write Newton’s second law.

$\sum F=ma$

Note that since the bus is on a hill, the acceleration due to gravity is pulling the bus backwards. In other words, the x-component of the force W is pulling the bus backwards.

$p-W\sin7^0=\dfrac{W}{g}a'$

Substitute the value of $p$ we found earlier.

$\dfrac{W}{g}a-W\sin7^0=\dfrac{W}{g}a'$

(multiply both sides of the equation by $\dfrac{1}{W}$, which eliminates W from our equation)

$\dfrac{a}{g}-\sin7^0=\dfrac{a'}{g}$

(Here, $a=3$ ft/$s^2$, and $g=32.2$ ft/$s^2$. $a'$ is the acceleration of the bus which slows it down while it is on the hill)

$\dfrac{3}{32.2}-\sin7^0=\dfrac{a'}{32.2}$

$a'=-0.9242$ ft/$s^2$

Again, the value we found is the acceleration which slows the bus down while it is climbing the hill. Since this is uniform, we can use the following equation to figure out the distance the bus travels to slow down to 50 mi/h.

$v^2=v_0^2+2a(x-x_0)$

(Where $v$ is final velocity, $v_0$ is initial velocity, $a$ is acceleration, $x$ is final displacement, and $x_0$ is initial displacement)

Let us first convert 60 mi/h and 50 mi/h to ft/s.

60 mi/h = 88 ft/s

50 mi/h = 73.3 ft/s

Substitute the values we know:

$v^2=v_0^2+2a(x-x_0)$

$73.3^2=88^2+(2)(-0.9242)(x-0)$

(Here, $x_0=0$ because we are assuming the bottom of the hill to be our origin)

$x=1282.8$ ft