Kinematics on an inclined plane

[JJBladester]

Homework Statement

In anticipation of a long 7° upgrade, a bus driver accelerates at a constant rate of 3ft/s² while still on a level section of the highway. Knowing that the speed of the bus is 60mi/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 50mi/h.

Answer: 0.242mi (1,278ft)

Homework Equations

g = 32.3ft/s²
θ = 7°
Initial acceleration = 3ft/s²
v_i = 60mi/hr = 88ft/s
v_f = 50mi/hr = 73.333ft/s
x_i = 0ft
x_f = ? ft

The Attempt at a Solution

$$\sum F_{x}=-mgsin\theta=ma$$

$$\sum F_{y}=N-mgcos\theta=0$$

Using the F_x equation, acceleration on the plane = -gsin(θ). The problem doesn't state anything about friction so my assumption is that friction doesn't play into the solution.

Acceleration starts at 3ft/s² at x_i and decreases by a constant -gsin(θ) as the bus travels up the grade.

I have a formula for acceleration when it depends on position but it produces the wrong answer.

$${v_{f}}^{2}={v_{i}}^{2}+\int_{0}^{x_{f}}a(x)dx$$

$$(73.\bar{3})^{2}-(88)^2=\int_{0}^{x_{f}}[3-gsin(\theta)]dx$$

$$(73.\bar{3})^{2}-(88)^2=(3-gsin(\theta))x_{f}$$

$$x_{f}=2565.6ft=.486mi\neq[the.correct.answer]$$

 

[Doc Al]

$${v_{f}}^{2}={v_{i}}^{2}+\int_{0}^{x_{f}}a(x)dx$$

You're off by a factor of 2 in that last term:
V²_f = V²_i + 2ax

 

[JJBladester]

Yeppers... That'll do it. And now the math works out and I have the correct answer. Thanks Doc!

 

[Patdon10]

i'm having trouble following the math involved. Can someone set me straight?

Vf^2 = V_0^2 + 2aD
73.3^2 - 80^2 = 2aD
-1022.27 = 2(3-9.81sin(7)x_f
-511.1355 = (3-9.81(sin7))x_f
-511.1355 = (3-1.1955)x_f
-511.1355 = 1.8044*x_f
x_f = -283.26 feet.

What am i missing?