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SaintJude

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The car is parked on a slope that makes an angle θ with the horizontal. If the handbrake is then released, and the car is allowed to roll down the slope, which of the following graphs represents the acceleration of the car versus the displacement?

The acceleration graph is just a straight horizontal line, b/c acceleration is constant.

Why would the beginning of the rolling not eve indicate acceleration?
 

SaCkO

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The car is parked on a slope that makes an angle θ with the horizontal. If the handbrake is then released, and the car is allowed to roll down the slope, which of the following graphs represents the acceleration of the car versus the displacement?

The acceleration graph is just a straight horizontal line, b/c acceleration is constant.

Why would the beginning of the rolling not eve indicate acceleration?
If it is horizontal it means that there is acceleration and that it has value.. Being horizontal means that it is constant. Since the forces acting on the car are constant and F=ma then acceleration stays constant too.. so as the car moves down and displacement increase (x axis), the acceleration stays the same on Y axis thus giving the graph a horizontal line.
 

MedPR

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The car is parked on a slope that makes an angle θ with the horizontal. If the handbrake is then released, and the car is allowed to roll down the slope, which of the following graphs represents the acceleration of the car versus the displacement?

The acceleration graph is just a straight horizontal line, b/c acceleration is constant.

Why would the beginning of the rolling not eve indicate acceleration?

The acceleration is gsintheta, only the velocity and displacement change.
 

pfaction

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Acceleration is a constant so it's a flat line as MedPR stated: gsinO flat across with time.
Velocity will be a proportional thing because it's at a constant rate due to acceleration being constant.
Displacement will be the right half of a parabola if you can see what I'm trying to say.
 
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