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Alex Ar [27]
3 years ago
6

Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim

e, another car is 1/2 mile east of the intersection, driving east (away from the intersection) at an unknown speed. The officer's radar gun indicates 25 mph when pointed at the other car (that is, the straight-line distance between the officer and the other car is increasing at a rate of 25 mph). What is the speed of the other car?
Physics
1 answer:
blagie [28]3 years ago
7 0

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

x=x_{0}+V t \\ x=\frac{1}{2}+V t

The distance between the police car and the intersection is,

y=y_{0}+V t

y=\frac{1}{2}-40 t

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })

z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)

The rate of change is,

2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)

2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots

Now finding z when t=0, from (1) we have

z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}

z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071

The officer's radar gun indicates 25 mph pointed at the other car then, \frac{d z}{d t}=25 when t=0, from

From (2) we get

2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)

2(0.7071)(25)=V+2 V^{2}(0)-40

35.36=V-40

V=35.36+40=75.36

Hence the speed of the car is 75.36 mph

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A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7
aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

f'=(\frac{v+v_r}{v+v_s})f

where

f is the original frequency

f is the apparent frequency

v is the velocity of the wave

v_r is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

v_s is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

v_s = v_A = -28.7 m/s (surface A is the source, which is moving towards the receiver)

v_r = +62.2 m/s (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz

(b) 0.232 m

The wavelength of a wave is given by

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m

(c) 1481.2 Hz

Again, we can use the same formula

f'=(\frac{v+v_r}{v+v_s})f

In the source frame (= on surface A), we have

v_s = v_B = -62.2 m/s (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

v_r = +28.7 m/s (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz

(d) 0.225 m

The wavelength of the wave is given by

\lambda=\frac{v}{f}

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m

8 0
3 years ago
A 15.5 kg mass vibrates in simple harmonic motion with a frequency of 9.73 Hz. It has a maximum displacement from equilibrium of
algol [13]

Answer:

12.14 cm

Explanation:

mass, m = 15.5 kg

frequency, f = 9.73 Hz

maximum amplitude, A = 14.6 cm

t = 1.25 s

The equation of the simple harmonic motion

y = A Sin ωt

y =  A Sin (2 x π x f x t)

put, t = 1.25 s, A = 14.6 cm, f = 9.73 Hz

y = 14.6 Sin ( 2 x 3.14 x 9.73 x 1.25)

y = 14.6 Sin 76.38

y = 12.14 cm

Thus, the displacement of the particle from the equilibrium position is 12.14 cm.

6 0
3 years ago
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