Aparticle 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?

s(4)=450

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

so

derivative it we get

so

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2