Question

A particle is moving with acceleration a(t)=30t+8. its position at time t=0 is s(0)=14 and its velocity at time t=0 is v(0)=13. What is its position at time t=4?

Answer 1

s(4)=450

Step-by-step explanation:

Given Data

a(t)=30t+8

at t=0 s(0)=14

at t=0 v(0)=13

s(4)=?

Solution

we know v is derivative of a

and s is derivative of v

a(t)=30t+8

derivative it we get

`v(t)=15t^(2)+8t+c\\ put t=0\\v(0)=15(0)^(2)+ 8(0)+c\\v(0)=c\\c=13`

so

`v(t)=15t^(2)+8t+13`

derivative it we get

`s(t)=(15)/(3)t^(3)+4t^(2)+13t+c\\    t=0\\s(0)=(15)/(3)(0)^(3)+4(0)^(2)+13(0)+c\\s(0)=c\\c=14`

so

`s(t)=(15)/(3)t^(3)+4t^(2)+13t+14\\t=4\\s(4)=(15)/(3)(4)^(3)+4(4)^(2)+13(4)+14\\s(4)=450`