Particle A starts from rest at x=0 with constant acceleration 2 m s⁻² at t=0. Particle B crosses x=0 at t=2 s and thereafter moves with constant speed 10 m s⁻¹ in the same direction. The time interval between the two meetings of A and B is:
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Question

Particle A starts from rest at x=0 with constant acceleration 2 m s⁻² at t=0. Particle B crosses x=0 at t=2 s and thereafter moves with constant speed 10 m s⁻¹ in the same direction. The time interval between the two meetings of A and B is:

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G1 Physics · Accepted Answer

For A: position of A = (1/2)(2)t² = t². For B, after t=2 s: position of B = 10(t−2). Meetings satisfy t² = 10t−20 ⇒ t²−10t+20=0 ⇒ t = 5 ± √5. Both roots are greater than 2 s. Therefore the time interval between meetings is (5+√5) − (5−√5) = 2√5 s.