Question:
A car is moving collinearly with a laser beam emitted by a transceiver. A laser pulse emitted at \( t = 0 \) s is received back by the transceiver 100 ns (nanoseconds) later after reflection from the car. A second pulse emitted at \( t = 0.1 \) s is received back 90 ns later. Given the speed of light is \( 3 \times 10^8 \) m/s, the average speed of the car in this interval is _________
A car is moving collinearly with a laser beam emitted by a transceiver. A laser pulse emitted at \( t = 0 \) s is received back by the transceiver 100 ns (nanoseconds) later after reflection from the car. A second pulse emitted at \( t = 0.1 \) s is received back 90 ns later. Given the speed of light is \( 3 \times 10^8 \) m/s, the average speed of the car in this interval is _________
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To find the speed of an object using time-of-flight measurements, use the time difference and speed of light to calculate the distance moved.
Updated On: Nov 25, 2025
- 54 kmph, moving towards the transceiver
- 108 kmph, moving towards the transceiver
- 54 kmph, moving away from the transceiver
- 108 kmph, moving away from the transceiver
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the problem.
The distance traveled by light is given by the formula \( d = c \cdot t \), where \( c \) is the speed of light and \( t \) is the time taken for the light to travel to and from the car. In the first case, the pulse takes 100 ns, and in the second case, it takes 90 ns. The difference in time corresponds to the relative motion of the car. Step 2: Calculating the distance.
The total round-trip distance traveled by light is: \[ \text{Distance} = c \times \text{time difference} \] Thus, the total distance for the second pulse is: \[ \text{Distance} = 3 \times 10^8 \times 90 \times 10^{-9} = 27 \, \text{m} \] Since this is the distance the car has moved, the average speed is: \[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{27}{0.1} = 54 \, \text{kmph} \] Step 3: Conclusion.
The car is moving at 54 kmph, and since the second pulse took less time, it indicates that the car is moving towards the transceiver. Thus, the correct answer is (A).
The distance traveled by light is given by the formula \( d = c \cdot t \), where \( c \) is the speed of light and \( t \) is the time taken for the light to travel to and from the car. In the first case, the pulse takes 100 ns, and in the second case, it takes 90 ns. The difference in time corresponds to the relative motion of the car. Step 2: Calculating the distance.
The total round-trip distance traveled by light is: \[ \text{Distance} = c \times \text{time difference} \] Thus, the total distance for the second pulse is: \[ \text{Distance} = 3 \times 10^8 \times 90 \times 10^{-9} = 27 \, \text{m} \] Since this is the distance the car has moved, the average speed is: \[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{27}{0.1} = 54 \, \text{kmph} \] Step 3: Conclusion.
The car is moving at 54 kmph, and since the second pulse took less time, it indicates that the car is moving towards the transceiver. Thus, the correct answer is (A).
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