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3 years ago

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Fiora starts riding her bike at 20 mi/h. After a while, she slows down to 14 mi/h, and maintains that speed for the rest of the

trip. The whole trip of 112 miles takes her 6.5 hours. For how long did she travel at 20 mi/h?

1 answer:

goblinko [34]3 years ago

4 0

Answer:

3.5 hours

Explanation:

Speed = distance/time

Let the distance that Fiora biked at 20 mi/h through be x miles and the time it took her to bike through that distance be t hours at 20 mi/h

Then, the rest of the distance that she biked at 14 mi/h is (112 - x) miles

And the time she spent biking at 14 mi/h the rest of the distance = (6.5 - t) hours

Her first biking speed = 20 mph = 20 miles/hour

Speed = distance/time

20 = x/t

x = 20 t (eqn 1)

Her second biking speed = 14 mph = 14 miles/hour

14 = (112 - x)/(6.5 - t)

112 - x = 14 (6.5 - t)

112 - x = 91 - 14t (eqn 2)

Substitute for x in (eqn 2)

112 - 20t = 91 - 14t

20t - 14t = 112 - 91

6t = 21

t = 3.5 hours

x = 20t = 20 × 3.5 = 70 miles.

(112 - x) = 112 - 70 = 42 miles

(6.5 - t) = 6.5 - 3.5 = 3 hours

Meaning that she travelled at 20 mi/h for 3.5 hours.

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<h3>Velocity</h3>

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Mass of the first ball: $100\; \rm g = 0.1\; \rm kg$ .
Mass of the second ball: $400\; \rm g = 0.4 \; \rm kg$ .

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Second ball: $p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}$ .
Sum: $10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1}$ given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be $30\; \rm kg \cdot m \cdot s^{-1}$ . The mass of the two balls, combined, is $0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg$ . Let the velocity of the two balls after the collision $v\; \rm m \cdot s^{-1}$ . (There's only one velocity because the collision had sticked the two balls to each other.)

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$v = 60$ .

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The kinetic energy of an object of mass $m$ and velocity $v$ is $\displaystyle \frac{1}{2}\, m \cdot v^{2}$ .

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Second ball: $\displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.4\; \rm kg \times \left(50\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J$ .
Sum: $500\; \rm J + 500\; \rm J = 1000\; \rm J$ .

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.

Mass of the two balls, combined: $0.5\; \rm kg$ .
Velocity of the two balls right after the collision: $60\; \rm m\cdot s^{-1}$ .

Sum of the kinetic energies of the two balls right after the collision:

$\displaystyle \frac{1}{2} \, m \cdot v^{2} = \frac{1}{2}\times 0.5\; \rm kg \times \left(60\; \rm m \cdot s^{-1}\right)^2 = 900\; \rm J$ .

Therefore, $1000\; \rm J - 900\; \rm J = 100\; \rm J$ of kinetic energy would be lost during this collision.

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