The graph shows the movement of a spacecraft which has fired its “reverse thrusters.” where will the ship be located relative to the origin at t = 25 s if it continues this constant acceleration? assume that the velocity of the spacecraft is zero at t = 0. at t = 2.5 s, position is exactly 7.00 m.

Let X(t)=the position function with respect to time, t.

We're given:

X(0)=12 (from graph)

v(0)=0

X(2.5)=7

There is enough information to construct X(t).

In general,

X(t)=kt^2+v(0)(t)+x0

where initial velocity=v(0)=velocity at time 0 = 0 (given)

x0=initial position at time 0 = 12 (from graph)

Substituting values, the position equation is

X(t)=kt^2+12

Substituting given value X(2.5)=7 => 7=k(2.5^2) + 12

Solve for k to get k=(12-7)/6.25 = -4/5

The position equation now becomes

X(t)=-4t^2/5+12

For t=25,

X(25)=-4(25^2)/5+12=-500+12 = 488 m

The vertex of the position curve x(t) is shown at x=12. To make the curve go through (t, x) = (2.5, 7), you need to subtract 5 from the vertex when t=2.5. (t/2.5) makes a factor that is 1 at t=2.5, and multiplying that factor by 5 gives the amount you need to subtract. Since it is a quadratic, we can square the (t/2.5) term to get the curve we're looking for. Hence x(t) = 12 - 5(t/2.5)^2.

Of course, when t=25, 25/2.5 = 10 and the square of that is 100, even for those who don't know their times tables. 5*100 = 500 and subtracting 12 from that gives 488--all in your head. All that math has a minus sign in front, so you end up with -488, the second selection.

x(t) = 12 - 5(t/2.5)²

x(25) = 12 - 5(25/2.5)²

... = -(5×10² - 12)

... = -488 . . . . meters