Question

The acceleration function (m/s^2) and the initial velocity are given for a particle moving along a line. Find each of the following:
a(t)=2t+3, v(0)=-4, 0 is less than or equal to t which is less than or equal to 3
A)the velocity at time t
B) The distance traveled during the given time interval.
C) Show that for motion in a straight line with constant acceleration a, and initial velocity v0, and initial displacement s0, the displacement

Answer 1

s'=v
v'=a

so to find v you integrate a

Answer 2

then to find the constant of integration there use v(0)=-4

Answer 3

is this a calculus based question?

Answer 4

the velocity is the area of the acceleration graph, if this is not calculus based

Answer 5

I was able to figure out parts a and b, I mostly need help with c

Answer 6

it's calculus based

Answer 7

your C part is missing things ...

Answer 8

let a(t) = a

integrate to find velocity

integrate again to fine position

Answer 9

oh whoops the full c says "Show that for motion in a straight line with constant acceleration a, initial velocity v0, and initial displacement s0, the displacement after time t is S=1/2at^2+v0t+s0

Answer 10

what do we get when we integrate a constant?
\[\int a~d=a\int dt = ??\]

Answer 11

... dropped a t at the start lol

Answer 12

you would get t right?

Answer 13

no wait i mean you would get ad

Answer 14

a (t+c) for some arbitraty constant c

v = at + ac, but ac is just still an arbitrary constant

v = at + c, what is c when t=0?

Answer 15

\[v =\int a~dt=at+c\]
\[v_0=a(0)+c\]
therefore
\[v = at+v_0\]
integrate it again ....

Answer 16

\[s=\int at+v_o~dt\]
\[s=\int at~dt+\int v_o~dt\]
\[s=a\int t~dt+v_0\int ~dt\]

Answer 17

thers no real ned to split it, just work it how you feel best