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Draw a velocity-time graph for the following data: Time (in seconds ) 2 4 6 8 10 12 14 Velocity (in m/s) 5 10 15 20 20 10 0 (a) What type of motion is represented during 0-6 seconds? (b) What is acceleration in the first 6 seconds? (c) Calculate the retardation (i.e negative acceleration) during 12-14 seconds. |
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Answer» (a) During the 0 -6 seconds the motion represented is uniform motion . (b) Acceleration in FIRST 6 seconds = a = (15-5) / (6 -2 ) = 10/4 = 2.5 m/s2 . (c) Distance TRAVELLED between the 6 to 8 seconds is , S=ut+12ft2=(15×2)+12×(20−152)×4=30+5=35 mS=ut+12ft2=15×2+12×20-152×4=30+5=35 m (d) Retardation = a' = ( 0-10)/2 = - 5 m/s2 |
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