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^{2}, where l is the contracted length.Now the observer reaches planet B, decelerates instantly to a complete stop and observes that planet A jumps back to distance x. Just before deceleration, he sees that the clock on planet A shows time t/γ, where t is the time on his clock. What happens after the deceleration? Does the time on clock A also jump to t or does it remain t/γ? And does the time on clock B, on the planet where he has landed remain (t-vx/c

^{2})/γ or does it jump forward to t?