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A train starts from rest and moves with constant acceleration 5m/s^2 for 10sec .Then moves uniformly for another 30sec and finally comes to rest in next 10sec .Find its average speed?

A train starts from rest and moves with constant acceleration 5m/s^2 for 10sec .Then moves uniformly for another 30sec and finally comes to rest in next 10sec .Find its average speed?

Grade:11

1 Answers

Tanay
19 Points
6 years ago
We know that,Average speed = (Total distance travelled)/(Total Time taken)Let`s divide the above question in three cases...Case 1: The train is acceleratingS(1) = ut+1/2at² = 0×10 + 1/2×5×10^2 = 250mV = u+at = 0+5×10 = 50m/sCase 2: The train is moving with constant speed...S(2) = (constant speed)× Time = 50×30 = 1500mCase 3: The train is retardingAcceleration (a)= (v-u)/t = (0-50)/10 = -5m/s²v²-u²=2aS(3)S(3)= (0-2500)/{2×(-5)} = 250mNow,Total distance travelled = S(1)+S(2)+S(3) = 250+1500+250 = 2000mTotal Time taken = 10+30+10 = 50secTherefore,Average velocity = 2000/50 = 40m/s

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