The Acceleration function (in m/s^2) and initial velocity for a particle moving along a line is given by a(t)= -14t-21, v(0)=28, 0

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irishboy123 · Answer

firstly, checking what you have done already:

$$a(t)= -14t-21$$
$$v(t)= -7t^2-21t + A$$
$$v(0) = 28 = A$$
$$v(t)= -7t^2-21t + 28$$

next, in terms of distance travelled, you integrate again:

$$x(t) = -\frac{7}{3}t^3-\frac{21}{2}t^2 + 28t + B$$

and in the first instance the answer might be x(3) - x(0) = ???  the B's cancel, nice.

BUT that will give you the difference between the positions or **displacements**  of x at t=3 and t=0, not necessarily the distance it has travelled in getting from one displacement to the other --- as the particle could change direction of travel.  note it is accelerating in the neg x direction during the whole period.

you need to look at the position where v(t) = 0, that is =>
$$t^2 + 3t - 4 = 0 ; \space{    } t = -4, 1$$
and then you need to add together the distances it travels in opposite directions during period 0≤t≤1 and 1≤t≤3, to get the total distance it has travelled.