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A train moving with a velocity of 88.1 km/hour North, increases its speed with a uniform acceleration of 0.250 m/s2 North until it reaches a velocity of 160.0 km/hour North. What distance did the train travel while it was increasing its velocity, in units of meters? Give the answer as a positive number |
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Answer» -INITIAL Velocity of Train = u = 88.1 Km/h = 24.4 ms-¹Final velocity of train = v = 160 Km/h = 44.4 ms-¹Acceleration of Train = a = 0.25 ms-²Using First Equation of MOTION,v - u = at44.4 - 24.4 = 0.25tt = 20/0.25 t = 80 s Again,DISTANCE traveled,S = ut + 1/2 at²S = (24.4)(80) + 1/2(0.25)(80)(80)S = 1952 + 800S = 2752 mFinding distance traveled USING THIRD equation of Motion,v² - u² = 2aS(44.4)² - (24.4)² = 2(0.25)S1971.36 - 595.36 = 0.5SS = 1376/0.5S = 2752 m Hence,Distance travelled = S = 2752 m |
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