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 after time t is
S=1/2at^2+v0t+s0

Im confused what to do for C

Answer 1

hint:
in order to find the space traveled, at time t, we have to integrate twice the acceleration a(t)

Answer 2

whereas in order to find the velocity, we have to integrate once the function a(t)

Answer 3

for example, the velocity v(t), is given by the subsequent computation:
\[\Large v\left( t \right) = \int {a\left( t \right)dt = } \int {\left( {2t + 3} \right)dt} = ...?\]

Answer 4

v(t)=t^2+3t-4 and then s(t)=t^3/3 +3/2t-4t

Answer 5

v(t) is right, whereas integrating once v(t) I get:
\[\Large s\left( t \right) = \frac{{{t^3}}}{3} + \frac{3}{2}{t^2} - 4t + k\]
where k is an arbitrary constant, whose meaning is the space at t=0, namely:
k=s(0)=s_0