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Laws of Motion topic -- Car being decelerated by a headwind

  • Thread starter Barclay
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  • #1
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Homework Statement


A car with mass 2500 kilograms travels along a straight level road at 90 km/h. A headwind begins to blow with a force of 200 N causing the car to slow down. Calculate the car’s deceleration.

Homework Equations


The book uses the formula F =ma so
upload_2014-12-30_22-1-27.png



3. Book's solution

upload_2014-12-30_22-1-27.png
= 0.08 m/s2


My confusion with the calculation comes from not knowing why the speed of the car is not considered. It’s a 2500 kg car driving at 90km/h so must have a force or a momentum that is important. Why is the car's speed not considered?


I need to ask a bit more about the question here please:

The second part of the question says

The car travelling at a steady speed collides with a stationary traffic cone of mass 10 kg. The traffic on accelerates in the direction of the car’s motion at 29 m/s2.
Calculate the force applied to the traffic cone by the car.


Book says: Calculation is F =ma = 10 x 29 = 290N

Book asks: What is the force applied by the cone to the car in the collision (hint in the paper is: Newton’s third law of motion).

Answer in the book is 290N. But shouldn’t the answer be -290N because “to every action is an equal and opposite reaction”.
 

Answers and Replies

  • #2
BvU
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Re deceleration: it is meant to be a simple exercise. No mention of cruise control either :)
so apply Newton 1
Re cone: Acceleration is in the direction of the car's motion, which (apparently) is chosen to be the positive direction (it says 29 m/s2, not -29 m/s2). Force applied by the car to the cone is then also considered positive. Force of cone on car is then -290 N.
 
  • #3
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For the first part of the question, the speed of the car that the question gives you is a red-herring; it tries to tricks students into believing that knowing the initial speed is important in this situation. Since you already have the the net force and the mass of the car, you can easily solve for the deceleration.
 
  • #4
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Force of cone on car is then -290 N.
So I'm correct ??
 
  • #5
BvU
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Yes. The book answer implicitly assumes only the magnitude of the force is asked for.
 
  • #6
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When a resultant force acts on an object, it's velocity always changes - whether it's a change in direction or magnitude. If there is no resultant force, then the object continues to be at rest or in a state of uniform motion (Newton's first law). The car's engine provides a constant force which counterbalances the frictional force and viscous force (air resistance) on the road, that's why the car keeps moving at 90 kmph, so you don't need to consider the speed at which the car moves when calculating the acceleration due to the unbalanced force exerted by the headwind. For the second part, if you're acquainted with momentum exchanges, then you'll see that the change in momentum of both the car and the cone is the same, and since they both stay in contact for the same time, the forces that the cone and the car experience are equal and opposite, so the magnitude of the force will be 290N. Notice that the question states the direction of the acceleration of the cone and then mentions the magnitude, so it's safe to assume that the answer is also expected to be in the form "290N in the direction opposite to the car's motion". So yes, if you just give a numerical answer, then it's -290N. If you give a sentence instead, it's what I mentioned in the previous line.
 

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