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2 years ago

A car accelerates uniformly from rest to 60 km/h in 30 s. What is its displacement during this time?

Engineering

1 answer:

Rina8888 [55]2 years ago

5 0

Answer:

252.62 m

Explanation:

Data provided :

Initial speed of the car, u = 0 Km/h (since, the car starts from the rest)

Final speed of the car, v = 60 km/h = 60 × (5/18) = 16.67 m/s

Time, t = 30 s

now,

from the Newton's equation of motion, we have

v = u + at

where,

a is the acceleration of the car

on substituting the values, we have

16.67 = 0 + a × 30

a = 0.55 m/s²

also,

v² - u² = 2as

where, s is the distance traveled

on substituting the values, we get

16.67² - 0² = 2 × 0.55 × s

s = 252.62 m

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anyanavicka [17]

Answer:

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Explanation:

To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.

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$X_C=\dfrac{1}{j\omega C}=\dfrac{-j}{100\pi\cdot 318.31\times10^{-6}F}\approx -j10.00000\,\Omega$

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Z1 = 8 +j4.99984 Ω

so the current is ...

I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°

The power factor is cos(-32.005°) ≈ 0.848 (lagging)

Branch 2

The impedance of branch 2 is ...

Z2 = 5 -j10 Ω

so the current is ...

I2 = 240/(5 +j10) ≈ 21.466∠63.435°

The power factor is cos(63.436°) ≈ 0.447 (leading)

Total current

The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.

It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)

It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°

The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)

The phasor diagram of the currents is attached.

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Additional comment

Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.

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The way to write this is

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