A. Velocity of the particle is v(t)=x′(t)=3t2−12t−15.
Hence v(t)=0⇒3t2−12t−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)=6t−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
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