CBSE Class 11-science Answered
Pls solve
Asked by pb_ckt | 16 Jun, 2019, 09:36: PM
Expert Answer
distance travelled = average velocity×time ...............(1)
average velocity = 72 km/hr = 20 m/s
distance travelled = 20×25 = 500 m .......................(2)
Let t be the time in seconds, car has travelled with uniform acceleration 5 m/s2
distance S1 travelled with uniform acceleration is obtained as, S1 = (1/2)×at2 = (1/2)×5×t2 = 2.5 t2 m ................(3)
velocity u after t seconds is given by, u = a×t = 5×t m/s
distance S2 travelled before coming to rest if retardation is 5 m/s2 :- S2 = u2/(2a) = 25t2 /(2×5) = 2.5 t2 .............(4)
Total distance S travelled is obtained by adding eqn.(3) and eqn.(4)
S = S1+S2 = 2.5t2 + 2.5t2 = 5 t2 = 500 m ..................(5)
Hence by solving eqn.(5) for t, we get t = 10 s ; Hence car has initially travelled 10 s with uniform acceleration .
Answered by Thiyagarajan K | 16 Jun, 2019, 10:30: PM
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