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Plzz Plzz answer this question..
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Answers
Answer:
Explanation:
ACCELERATION : 3 T²+4 T-2
(1) INTEGRATING ACCELERATION GIVES VELOCITY ;
∫A=∫3 T²+4 T-2
V = (3 T³/3) + (4 T²/2) - 2 T
V = T³+ 2 T²-2 T
VELOCITY AFTER 2 SEC
V = 2³+2*2²-2*2
V = 8+8-4
V = 16-4
V = 12 M/S
(2) V₁ = 1³+2*1²-2*1
V₁ = 1+2-2
V₁= 1 M/S.............(1)
V₃= 3³+2*3²-2*3
V₃= 27+18-6
V₃=39 M/S..................(2)
CHANGE IN VELOCITY = V₃-V₁
= 39-1
= 38 M/S
(3) INTEGRATING VELOCITY GIVES DISTANCE TRAVELLED
∫V = ∫T³+2 T²-2 T
S =( T⁴/4) +(2 T³/3) -(2 T²/2)......................[A]
S₂=( 2⁴/4)+(2*2³/3)-(2*2²/2)
S₂= 4 +16/3-4
S₂= 16/3 M.................(1)
S₄= (4⁴/4)+(2*4³/3)-(2*4²/2)
S₄=64+128/3-16
S₄=48+128/3
S₄=272/3................(2)
DISTANCE TRAVELLED = S₄-S₂
= 272/3-16/3
= 256/3
= 85.33 M
(4)LOCATION INITIALLY IMPLIES S AT T=0
WHEN WE SUBSTITUTE T=0 AT[A]
S = 0 M
THAT IS BODY IS AT ORIGIN.
HOPE THIS HELPS.