The particle's motion is described by the equation d = t2 - 8t + 15 where d and t and measured in metres and seconds. Show that the particle is at rest when t = 4.

Is there any value in factoring the equation into (t - 5)(t - 3) and if so then what?


Do we set the lim as t approaches 4 and calculate

f(4 + h) - f(4) all divided by h

=lim as t approaches 4 (4 + h)^2 - 8(4 + h) + 15 - (4)^2 -8(4) + 15 all divided by h

The particle is at rest when the derivative, or slope, is 0. Because the derivative of the position function is velocity. The particle is at rest when the velocity is 0.

You can find the derivative, but no need to use first principles unless you were told to.

d=t^{2}-8t+15

\frac{d}{dt}[t^{2}-8t+15]=2t-8

Solve

2t-8=0 for t.