Kinematics on an inclined plane

In summary, a bus driver accelerates at a constant rate of 3ft/s2 while on a level section of the highway, and then continues at the same rate while climbing a 7° grade. The driver does not change the throttle or shift gears. The distance traveled by the bus when its speed decreases from 60mi/h to 50mi/h is approximately 0.242mi (1,278ft). The equations used include g = 32.3ft/s2, θ = 7°, vi = 60mi/hr (88ft/s), vf = 50mi/hr (73.333ft/s), and xf = ?ft. The correct equation for calculating distance is Vf^2 =
  • #1
JJBladester
Gold Member
286
2

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
vi = 60mi/hr = 88ft/s
vf = 50mi/hr = 73.333ft/s
xi = 0ft
xf = ? ft

The Attempt at a Solution



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

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

Using the Fx 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 xi 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]
 
Last edited:
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  • #2
JJBladester said:
[tex]{v_{f}}^{2}={v_{i}}^{2}+\int_{0}^{x_{f}}a(x)dx[/tex]
You're off by a factor of 2 in that last term:
V2f = V2i + 2ax
 
  • #3
Yeppers... That'll do it. And now the math works out and I have the correct answer. Thanks Doc!
 
  • #4
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?
 
  • #5


As a scientist, it is important to always double check our assumptions and equations to ensure that our solution is accurate. In this case, it seems that the formula you used for acceleration may not be applicable in this scenario since the acceleration is not constant. It is important to consider the varying acceleration due to the incline of the plane.

To solve this problem, we can use the equations of motion for constant acceleration, where the acceleration is not necessarily constant but varies with position. In this case, we can use the equation vf^2=vi^2+2ad, where vf is the final velocity, vi is the initial velocity, a is the average acceleration, and d is the distance traveled.

In this problem, we can find the average acceleration by taking the average of the initial and final accelerations. So, a_avg = (3+ (-gsin(θ)))/2.

Next, we can plug in the given values to the equation vf^2=vi^2+2ad and solve for d. This will give us the distance traveled by the bus up the grade when its speed has decreased to 50mi/h.

d = [vf^2-vi^2]/2a_avg

= [(73.333)^2 - (88)^2]/[2(1.5 - (32.3sin(7)))/2]

= 1278 ft = 0.242 mi

Therefore, the distance traveled by the bus up the grade when its speed has decreased to 50mi/h is 0.242 mi or 1278 ft.
 

Related to Kinematics on an inclined plane

1. What is kinematics on an inclined plane?

Kinematics on an inclined plane is the study of the motion of objects on a surface that is not flat, but instead tilted at an angle. It involves analyzing the various forces acting on an object and how they affect its motion.

2. What are the key concepts in kinematics on an inclined plane?

The key concepts in kinematics on an inclined plane include gravity, normal force, friction, and the angle of inclination. These factors all play a role in determining the acceleration and velocity of an object on the inclined plane.

3. How do you calculate the acceleration of an object on an inclined plane?

The acceleration of an object on an inclined plane can be calculated using the formula a = gsinθ, where a is the acceleration, g is the acceleration due to gravity (9.8 m/s²), and θ is the angle of inclination.

4. How does the angle of inclination affect the motion of an object on an inclined plane?

The angle of inclination has a significant impact on the motion of an object on an inclined plane. A steeper angle will result in a greater acceleration and a faster velocity, while a shallower angle will result in a slower acceleration and lower velocity.

5. What is the difference between kinetic and potential energy on an inclined plane?

Kinetic energy is the energy an object possesses due to its motion, while potential energy is the energy an object has due to its position or height. On an inclined plane, an object's potential energy decreases as it moves down the incline, while its kinetic energy increases.

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