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lara31 [8.8K]
3 years ago
6

A train starting from rest picks up a speed of 20 m/s in 20 s while travelling on a straight path. It continues to move at the s

ame speed for the next 50 s. It is then brought to rest in the next 10 s.
a. Plot the speed-time graph for the train..

e. How much distance did the train cover during this entire journey?

Physics
1 answer:
Andrej [43]3 years ago
8 0

Answer:

1300 m

Explanation:

As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from 20 s to 70 s is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.

The total distance covered by the train during the entire journey is the area of the speed-time graph.

Area=\frac{1}{2}(20\times 20)+ 50\times 20+\frac{1}{2}(20\times 10)

=200+1000+100

=1300

As velocity is in m/s and time is in s so the unit of area is m

Hence, the total distance is  1300m.

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A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a
Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

Z_{R} = R

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ω = voltage source angular frequency, L = inductance

Z_{C} = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

Z_{R} = 217Ω

Z_{L} = j(220)(0.875) = j192.5Ω

Z_{C} = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = Z_{R} + Z_{L} + Z_{C}

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

v_{R} = I×Z_{R}

v_{L} = I×Z_{L}

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Given values:

I = (0.0569∠1.147+220t)A

Z_{R} = 217Ω = (217∠0)Ω

Z_{L} = j192.5Ω = (192.5∠π/2)Ω

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Plug in and calculate each component's voltage:

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v_{C} = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

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Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

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v_{C} = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

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