Question

The position of a particle which moves along a straight line is defined by the relation x=t^3-6t^2-15t+40, where x is expressed in feet and t in second.A. determine the time at which the velocity will be zero B.the position and distance traveled by particle at that time C.the acceleration of the particle at that time D.the distance traveled by the particle from t=4s to t=6s.

Answer 1

A. Velocity of the particle is $v(t) = x'(t) = 3 t^2 - 12 t - 15$.

Hence $v(t) = 0 \Rightarrow 3 t^2 - 12 t - 15 = 0$, which has positive solution $t = 5$, so velocity is zero at $t = 5$.

B. Substituting $t = 5$ into $x(t)$, obtain $x(5) = -60$. The distance is $|x(5) - x(0)| = |-60 - 40| = 100$.

C. Acceleration is $a(t) = v'(t) = 6 t - 12$, hence $a(5) = 18$.

D. $d(4, 6) = d(4,5) + d(5,6)$, since the body has stopped at $t=5$, and started moving into opposite direction.

$d(4,5) = |-52 - (-60)| = 8$,

$d(5,6) = |-50 - (-60)| = 10$,

hence $d(4,6) = 8 + 10 = 18$