Question

if the relation between acceleration and Tiime for an object is given as a=2t +4t^2 then calculate the position of object from the origin at t=4s

Answer 1

GIVEN:-

a=2t +4t^2
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we know that ...

$a = \frac{dv}{dt}$
=> v= ∫ dv. = ∫ adt. = ∫ (2t+4t^2)dt

=> v= t^2 + (4/3)t^3
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we know that...

$v = \frac{dx}{dt}$
=> x=∫ dx = ∫ vdt

=> x= ∫ [t^2 + (4/3) t^3]dt

=> x= (t^3)/3 + (t^4)/3
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so...

x =?? at t=4s

=> x= (4^3)/3 + (4^4)/3

=> x= (1/3){4^3+4^4}

=> x= (1/3){64+256}

=>x=320/3

=> x= 106.667m
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hence, the position of object from origin is 106.667m
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hope it helps you..

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