PHYSICS HIGH SCHOOL

Sally travels by car from one city to another. She drives for 30.0 min at 80.0 km/hr, 12.0 min at 105.0 km/hr, and 45.0 min at 40.0 km/hr, and she spends 15.0 min eating lunch and buying gas. A. Determine the average speed for the trip.
B. Determine the total distance traveled.

Answers

Answer 1

Answer:

Answer:

A) The average speed for the trip is $v_(average) =53,53(km)/(h)$ .

B) The total distance traveled is $d_(total)=91\ km$ .

Explanation:

Knowing that 1 hour has 60 minutes, we have that for each part of the trip:

$v_(1)=80(km)/(h)\ t_(1)=0,5\ h$

$v_(2)=105(km)/(h)\ t_(2)=0,2\ h$

$v_(3)=40(km)/(h)\ t_(3)=0,75\ h$

$v_(4)=0(km)/(h)\ t_(4)=0,25\ h$

Total distance traveled

The distance traveled is the speed multiplied by a period of time. The total distance traveled is calculated adding up the speeds multiplied by the given periods of time.

$d_(total)=d_(1)+d_(2)+d_(3)$

$d_(total)=v_(1).t_(1)+v_(2).t_(2)+v_(3).t_(3)$

$d_(total)=80(km)/(h).0,5\ h+105(km)/(h).0,2\ h+40(km)/(h).0,75\ h$

$d_(total)=40\ km+21\ km+30\ km$

$d_(total)=91\ km$

B) The total distance traveled is $d_(total)=91\ km$ .

Average speed

The average speed could be calculated using the total distance traveled $d_(total)$ divided by the total time $t_(total)$ . We need to take into account the time spent eating lunch and buying gas.

$v_(average)=(d_(total))/(t_(total))$

$v_(average)=(v_(1).t_(1)+v_(2).t_(2)+v_(3).t_(3)+v_(4).t_(4))/(t_(1)+t_(2)+t_(3)+t_(4))$ $v_(average)=(91\ km)/(0,5\ h+0,2\ h+0,75\ h+0,25\ h)$

$v_(average)=(91\ km)/(1,70\ h)$

$v_(average)=53,53(km)/(h)$

A) The average speed for the trip is $v_(average) =53,53(km)/(h)$ .

Answer 2

Answer: Total distance = (30/60 x 80) + (12/60 x 105) + (45/60 x 40) = 0.5 x 80 + 0.2 x 105 + 0.75 x 40 = 40 + 21 + 30 = 91 km

Average distance = total distance / time taken = 91 / (30/60 + 12/60 + 45/60) = 91/ (0.5 + 0.2 + 0.75) = 91/1.45 = 62.76 km/hr

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Answers

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Explanation:

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Answers

The displacement vector, d, can be recomposed as follows: d = (Dx, Dy) and the magnitude of the displacement vector is approximately 11.24 meters.

To recompose the displacement vector, d, using the given X and Y components, you can use vector addition. The X and Y components represent the horizontal and vertical displacements, respectively.

Given:

Dx = 6.68 m (left)

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The displacement vector, d, can be recomposed as follows:

d = (Dx, Dy)

So, the displacement vector is (6.68 m left, 9.04 m down). This means that the object moved 6.68 meters to the left (in the negative X direction) and 9.04 meters downward (in the negative Y direction) from its starting point.

The magnitude of the displacement can be found using the Pythagorean theorem:

|d| = √((Dx)^2 + (Dy)^2)

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|d| ≈ √(126.21 m^2)

|d| ≈ 11.24 meters

So, the magnitude of the displacement vector is approximately 11.24 meters.

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