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A patrol car is moving at a constant speed towards a stationary observer - NSC Physical Sciences - Question 6 - 2019 - Paper 1

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A patrol car is moving at a constant speed towards a stationary observer. The driver switches on the siren of the car when it is 300 m away from the observer. The o... show full transcript

Worked Solution & Example Answer:A patrol car is moving at a constant speed towards a stationary observer - NSC Physical Sciences - Question 6 - 2019 - Paper 1

Step 1

6.1.1 Calculate the speed of the patrol car.

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Answer

To calculate the speed of the patrol car, we can use the formula for speed:

v=dtv = \frac{d}{t}

where:

  • d=300d = 300 m (the distance to the observer)
  • t=10t = 10 s (the time taken to reach the observer).

Substituting the values:

v=300 m10 s=30 m s1v = \frac{300 \text{ m}}{10 \text{ s}} = 30 \text{ m s}^{-1}

Thus, the speed of the patrol car is 30 m s⁻¹.

Step 2

6.1.2 State the Doppler effect.

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Answer

The Doppler effect refers to the change in frequency (or pitch) of a wave in relation to an observer who is moving relative to the wave source. When the source approaches the observer, the frequency increases, and when it moves away, the frequency decreases.

Step 3

6.1.3 The detected frequency suddenly changes at t = 10 s. Give a reason for this change.

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Answer

The detected frequency changes at t = 10 s because this is the moment when the patrol car passes the observer. At this point, the source of the sound (the siren) is moving directly away from the observer, resulting in a decrease in the detected frequency.

Step 4

6.1.4 Calculate the frequency of the sound emitted by the siren.

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Answer

We can use the Doppler effect formula to find the frequency of the sound emitted by the siren:

ft=v+vovvsfsf_t = \frac{v + v_o}{v - v_s} f_s

For a stationary observer ( v_o = 0), the equation simplifies to:

ft=vvvsfsf_t = \frac{v}{v - v_s} f_s

At the point where the detected frequency is 932 Hz when approaching, and v=340v = 340 m/s. We can rearrange to find fsf_s:

932=34034030fs932 = \frac{340}{340 - 30} f_s

Solving gives:

fs=932×34030340=849.76 Hz.f_s = 932 \times \frac{340 - 30}{340} = 849.76 \text{ Hz}.

Thus, the frequency of the sound emitted by the siren is approximately 849.76 Hz.

Step 5

6.2 State TWO applications of the Doppler effect.

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Answer

  1. Radar and sonar systems are often used to determine the speed of objects, such as vehicles or fish, by analyzing the frequency changes of reflected waves.

  2. The Doppler effect can also be utilized in astronomy to determine whether stars or galaxies are moving towards or away from Earth, by observing changes in the frequency of light from those objects.

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