starting from rest a car moves with uniform acceleration and attains a velocity of 10m/s in 10 s. it then moves with uniform speed for 15s and is then brought to rest in 12s under uniform retardation. find the total distance covered using velocity-time graph
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Let for our convince take 3 parts of displacements.
In the first part displacement = 1/2×a×t^2
=1/2× (v/t)×10×10
=1/2×(10/10)×10×10
=50m.
In the second part displacement=v×t
=10×15=150m.
In the third part displacement=1/2×a×t^2
=1/2×(v/t)× t^2 =1/2×(10/12)×12×12
=120/2
=60m.
Total displacement =50+150+60
=260.
In the first part displacement = 1/2×a×t^2
=1/2× (v/t)×10×10
=1/2×(10/10)×10×10
=50m.
In the second part displacement=v×t
=10×15=150m.
In the third part displacement=1/2×a×t^2
=1/2×(v/t)× t^2 =1/2×(10/12)×12×12
=120/2
=60m.
Total displacement =50+150+60
=260.
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