Question

A particle is moving with acceleration a ( t ) = 18 t + 4 . its position at time t = 0 is s ( 0 ) = 8 and its velocity at time t = 0 is v ( 0 ) = 7 . What is its position at time t = 5 ?

Answer 1

s = 468

Step-by-step explanation:

Answer 2

Given:

a(t) = 18t + 4

at t = 0 :

s(0) = 8

v(0) = 7

where,

s = position of particle

v = velocity of particle

Now, we know the following relations:

a =
`(dv)/(dt)`

v = ∫a dt = ∫(18t + 4)dt = 9
`t^(2)` + 4t + C

at t = 0 :

v(0) = 9(0) +4(0) + C

⇒ C = v(0) = 7

⇒ v (t) = 9
`t^(2)` + 4t +7

Now, using the relation:

v =
`(ds)/(dt)`

s = ∫v dt = ∫(9
`t^(2)` + 4t +7)dt = 3
`t^(3) + 2t^(2)` + 7t + C

at t = 0 :

s(0) = 3(0) + 2(0) + 7(0) + C

⇒ C = s(0) = 8

⇒ s(t) = 3
`t^(3) + 2t^(2)` + 7t + 8

Now, position at t = 5 :

s(5) = 3
`(5)^(3) + 2(5)^(2)` + 7(5) + 8

s = 468