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CBSE Class 11-science Answered

Points a and b are separated by 1 km a started moving towards b with a constant acceleration of5 m/s^2 while b starts moving away from a. With a constant speed of 1 m/s find -1. At what point will they meet 2. After what time will they meet 3. What is the velocity of AandB when they meet

Asked by aarushsingh3117 | 05 Jun, 2019, 08:41: AM

Expert Answer

Let A meet B at after t seconds.

Distance travelled by A = (1/2)at² = (1/2)×5×t² ..........................(1)

Distance travelled by B = v×t = 1×t ..................................(2)

where a is acceleration of A and v is uniform speed of B.

since initially A and B are 1 km apart, from eqn.(1) and eqn.(2), we have, (5/2)t² - t = 1000 ......................(3)

By solving eqn.(3), we get t = 20.2 s

(1) at what point they will meet :- distance travelled by A = (1/2)×5×20.2×20.2 = 1020.1 m

Hence both A and B will meet at a distance 1020.1 m from starting point of A

(2) after what time they will meet , t = 20.2 s

(3) velocity of A at meeting point = a×t = 5×20.2 = 101 m/s ; Since B is moving with constant speed, velocity of B = 1 m/s

Answered by Thiyagarajan K | 05 Jun, 2019, 10:13: AM

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