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SI unit of a bus starting from rest moves with uniform acceleration of 0.2 m/ s^2, for 300s. calculate the speed acquited and distance moved |
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Answer» question, we are provided with the values and measures of physical quantities involved in Kinematics (Study of MOTION in physics).GiveN:Starts from restAcceleration of the bus = 0.2 m/s²Time taken = 300 seconds.To find the FINAL velocity and distance covered....?This can be done by using the three equations of motion which are:V = u + ats = ut + 1/2 at²v² = u² + 2asSo, LET's solve by using these equations,By using first equation of motion,a = 0.2 m/s²u = 0 m/s (Initially at REST)t = 300 s➝ v = u + atPlug in the given values,➝ v = 0 + 0.2(300)➝ v = 0 + 60 m/s➝ v = 60 m/sHence the final velocity of the body is 60 m/s.━━━━━━━━━━━━━━━━━━━━Now we have to find the distance covered by the body. For this we can use the 3rd equation of motion.v = 60 m/su = 0 m/sa = 0.2 m/s²➝ v² = u² + 2asPlugging the given values,➝ 60² = 0² + 2(0.2)s➝ 3600 = 0.4s➝ s = 36000/4 m➝ s = 9000 mHence the distance covered by the bus = 9000 mAnd we are done !!! |
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