(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

In anticipation of a long 7° upgrade, a bus driver accelerates at a constant rate of 3ft/s^{2}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)

2. Relevant equations

g = 32.3ft/s^{2}

θ = 7°

Initial acceleration = 3ft/s^{2}

v_{i}= 60mi/hr = 88ft/s

v_{f}= 50mi/hr = 73.333ft/s

x_{i}= 0ft

x_{f}= ??? ft

3. The attempt at a solution

[tex]\sum F_{x}=-mgsin\theta=ma[/tex]

[tex]\sum F_{y}=N-mgcos\theta=0[/tex]

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/s^{2}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.

[tex]{v_{f}}^{2}={v_{i}}^{2}+\int_{0}^{x_{f}}a(x)dx[/tex]

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

[tex](73.\bar{3})^{2}-(88)^2=(3-gsin(\theta))x_{f}[/tex]

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

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# Homework Help: Kinematics on an inclined plane

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