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NikAS [45]
3 years ago
7

You normally drive a 12-h trip at an average speed of 100 km/h . Today you are in a hurry. During the first two-thirds of the di

stance, you drive at 116 km/h
If the trip still takes 12 h, what is your average speed in the last third of the distance?
Physics
1 answer:
kherson [118]3 years ago
8 0

Answer:

78 km/h

Explanation:

If I normally drive a 12 hour trip at an average speed of 100 km/h, my destination has a total distance of:

  • 100 km/h · 12 h = 1,200 km

Today, I drive the first 2/3 of the distance at 116 km/h. Let's first calculate what 2/3 of the normal distance is.

  • 1,200 * 2/3 = 800 km

I've driven 800 km already. I need to drive 400 km more to reach my final destination. I need to figure out my average speed during this last 1/3 of the distance.

To do this, I first need to calculate how much time I spent driving 116 km/h for the past 800 km.

  • 116 km/1 h = 800 km/? h
  • 800 = 116 · ?
  • ? = 800/116
  • ? = 6.89655172

I spent 6.89655172 hours driving during the first 2/3 of the distance.

Now, I need to subtract this value from 12 hours to find the remaining time I have left.

  • 12 h - 6.89655172 h = 5.10344828 h  

Using this remaining time and my remaining distance, I can calculate my average speed.

  • ? km/1 hr = 400 km/5.10344828 h
  • 5.10344828 · ? = 400
  • ? = 400/5.10344828
  • ? = 78.3783783148  

My average speed during the last third of the distance is around 78 km/h.

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  • Magnitude: 12.1 N.
  • Direction: 17.0° to the 8 N force.
<h3>Explanation</h3>

Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.

  • \displaystyle F_{\text{1, Parallel}} = F_1 \cdot \cos{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}.
  • \displaystyle F_{\text{1, Normal}} = F_1 \cdot \sin{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}.

The sum of forces on each direction will be the resultant force on that direction:

  • Resultant force parallel to the 8 N force: (8 + \dfrac{5\sqrt{2}}{2})\;\text{N}.
  • Resultant force normal to the 8 N force: \dfrac{5\sqrt{2}}{2}\;\text{N}.

Apply the Pythagorean Theorem to find the magnitude of the resultant force.

\displaystyle \Sigma F = \sqrt{{(8 + \dfrac{5\sqrt{2}}{2})}^2 + {(\dfrac{5\sqrt{2}}{2})}^2} = 12.1\;\text{N} (3 sig. fig.).

The size of the angle between the resultant force and the 8 N force can be found from the tangent value of the angle. Tangent of the angle:

\displaystyle \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}} = \dfrac{8 + \dfrac{5\sqrt{2}}{2}}{\dfrac{5\sqrt{2}}{2}} \approx 0.306491.

Find the size of the angle using inverse tangent:

\displaystyle \arctan{ \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}}} = \arctan{0.306491} = 17.0\textdegree.

In other words, the resultant force is 17.0° relative to the 8 N force.

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