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3 years ago

10

Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a

ramp at a high speed v0 and travels 690 ft in 8.5 s at constant deceleration before its speed is reduced to v0/2. Assume the same constant deceleration.a) Determine the additional time required for the truck to stop. The additional time required for the truck to stop is ______s b) the additional distance traveled by the truck is ______ft

1 answer:

Virty [35]3 years ago

5 0

Answer:

a) The additional time required for the truck to stop is <u>8.5 seconds</u>

b) The additional distance traveled by the truck is <u>230.05 ft</u>

Explanation:

Since the acceleration is constant, the average speed is:

(final speed - initial speed) / 2 = 0.75 v0

Since travelling at this speed for 8.5 seconds causes the vehicle to travel 690 ft, we can solve for v0:

0.75v0 * 8.5 = 690

v0 = 108.24 ft/s

The speed after 8.5 seconds is: 108.24 / 2 = 54.12 ft/s

We can now use the following equation to solve for acceleration:

$v^2 - u^2 = 2*a*s$

$54.12^2 - 108.24^2 = 2*a*690$

a = -6.367 m/s^2

Additional time taken to decelerate: 54.12/6.367 = 8.5 seconds

Total distance traveled:

$v^2 - u^2 = 2*a*s$

0 - 108.24^2 = 2 * (-6.367) * s

solving for s we get total distance traveled = 920.05 ft

Additional Distance Traveled: 920.05 - 690 = 230.05 ft

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A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to

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Answer:

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b) V_1 = V_2 = (16/3) * 10^2 V

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Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

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a) the original charge of the system,

(b) the ﬁnal potential difference across each capacitor

(c) the ﬁnal energy of the system

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Solution:

a)

- The initial charge in the circuit is the one carried by the first charged capacitor.

Q_initial = C_1*V

Q_initial = 20*10^-6 * 800

Q_initial = 16 * 10^-3 C

b)

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

Q_2 = Q_1*C_2/C_1

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- using conservation of charge:

Q_initial = 1.5*Q_1

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V_2 = V_1 = Q_1 / C_1

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c)

- The energy in the system is:

E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

E = 64/15 J

d)

- The decrease in energy of the capacitors is:

dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

dE = 0.5*10^-6*20*800^2 - (64/15)

dE = 32/5 - 64/15 = 32/15 J

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Answer:

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