GATE 2022 IN Q.39: solution and answer

GATE 2022 · INQ.39 · 2 marks · MCQ

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 × 10^8 m/s, the average speed of the car in this interval is ________

(A)
54 kmph, moving towards the transceiver
(B)
108 kmph, moving towards the transceiver
(C)
54 kmph, moving away from the transceiver
(D)
108 kmph, moving away from the transceiver

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Answer and solution

ANSWER

The car moves towards the transceiver at 54 km/h, option (A).

Official answer: ASource: Official answer key, GATE 2022 IN (IIT Kharagpur)

Time-of-flight ranging: the pulse travels to the car and back, so the one-way distance is $d = \frac{c t _{round}}{2}$ .
First pulse: $d_{1} = \frac{3 \times 1 0 ^{8} m/s \times 100 \times 1 0 ^{- 9} s}{2} = 15 m$ .
Second pulse: $d_{2} = \frac{3 \times 1 0 ^{8} m/s \times 90 \times 1 0 ^{- 9} s}{2} = 13.5 m$ .
The distance has decreased ( $d_{2} < d_{1}$ ), so the car is moving towards the transceiver. The change is $Δ d = 15 - 13.5 = 1.5 m$ .
The pulses are emitted $Δ t = 0.1 s$ apart, so the average speed is $v = \frac{1.5 m}{0.1 s} = 15 m/s$ .
Convert: $v = 15 \times 3.6 = 54 km/h$ , directed towards the transceiver. This is option (A).

KEY IDEA

A laser pulse's round-trip time gives the distance as $d = c t /2$ . The change in distance between two pulses, divided by the time between emissions, gives the car's speed. A shrinking distance means the car is approaching.

WATCH OUT FOR

Forgetting the factor of 2 (the pulse covers the path twice) gives 108 km/h, which is option B or D. Another slip is using the 100 ns or 90 ns round-trip time as the time interval for the speed, instead of the 0.1 s between emissions.

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GATE 2022 IN Q.39: solution and answer

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