Difficult calculus problem. Dare an attempt to solve?

Question

If the acceleration a(t) = 30 for all values of t, find a corresponding position function for which s(0) = 0 and v(0) = 40, all in appropriate units of measure. s(t) =

possible book solutions...

a. -10t^2 + 40t
b. -5t^2 + 40t
c. 5t^2 + 40t
d. 10t^2 + 40t
e. 15t^2 + 40t
f. none of these

Answer 1

d^2/dt^2=30
integrating
ds/dt=30t+C1
=30t+40 as V0=40,given
integrating again
s=30t^2/2+40t+C2
=15t^2+40t as s0=0 ,given
choice e

Answer 2

Difficult Calculus Problem

Answer 3

Answer is e.

s(t)=integration of v(t)
v(t)=integration of a(t)

v(t)=integration of 30
= 30t+constant
since v(0)=40
v(t)=30t+40
s(t)=integration of v(t)
=integration of 30t+40
=15t^2 +40t
Therefore answer is e

Answer 4

v(t) = v(0) + at
s(t) = s(0) + v(0) t + (1/2)at^2
s(t) = 0 + 40 t + 15 t^2


Ana

Answer 5

e. 15t^2 + 40t

Answer 6

15t² + 40t


Doug

Answer 7

the dazzling answer is -one million / ln(x) + C with out genuinely the fee enclosures. and that i see that Luiz already confirmed you the thank you to do the combination by using factors. verify... d/dx {-one million/ln(x)} = - d/dx {one million/ln(x)} = - d/dx {ln x}^(-one million) = - (-one million) (ln x)^(-2) d/dx {ln x} = [ one million / (ln x)^2 ] (one million/x) = [ x (ln x)^2 ]^(-one million)