Question

A mass is moving in the positive x direction with constant acceleration a(t)=-9.8. Find v(t) if v(0)=3 and Find s(t) if v(0)=3 and s(0)=4. For v(t) I got -4.9t^(2) + C. But for the second part, I am confused.

Answer 1

these are the notes I found. v'(t)=a(t), v(0)=initial velocity
and
s'(t)=v(t), s(0)= initial position

Answer 2

yes we have a=v'
so we need to integrate a(t)=-9.8

Answer 3

how did you find v(t)=-4.9t+c?

Answer 4

it is correct but do you know the method

Answer 5

to find the constant c
we know that v(0)=3
so v(0)=c=3
so v(t)=-4.9t+3
the method you used to get v is the same one to get s(t)
where v(t)=s'(t)

Answer 6

oh hold on that's wrong
v(t)=-9.8t+c not t^2
so c=3
v(t)=-9.8t+3

Answer 7

t^2 will be in the position function not v
it will be s(t)

Answer 8

oh okay thank you for correcting my first part

Answer 9

as for the second, let me see...I think i still need help though

Answer 10

basically s'(t) will be a constant, yeah?

Answer 11

correction in instruction s(0)=4

Answer 12

is s(t) then just equal to 4?