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A sprinter runs a 100 m race. The sprinter has a constant acceleration from rest of 2.5ms-2 until reaching a speed of 10 ms. The speed then remains constant until the end of the race. Which time does it take the sprinter to run the race? B 10 s с 12 s D A 8.9 s 14 s |
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Answer» Answer: C.The motion consists of 2 parts:1. motion with constant acceleration of 2.5 m s-22. motion with constant speed of 10 m s-1Let the duration of the acceleration be t1.v = u + atTime t1 = (v – u) / a = (10 – 0) / 2.5 = 4 sLet the distance travelled during the acceleration be s1.Acceleration a = 2.5 m s-2Initial speed u = 0 (at start)Final speed v = 10 m s-1We want to find distance s1. The equation that relates a, v, u and s is v2 = u2 + 2asDistance s = (v2 – u2) / 2aDistance travelled s1 = (v2 - u2) / 2a = (102 – 0) / (2 × 2.5) = 20 MTHE remaining distance of the 100 m race is COVERED at the constant speed of 10 m s-1.Remaining distance = 100 – 20 = 80 mLet the time taken to cover the remaining distance be T2.Speed = distance / timeTime t2 = distance / speed = 80 / 10 = 8 sTotal time taken = 8 + 4 = 12 s |
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