University Calculus: Early Transcendentals (3rd Edition)

a) The particle moves forward from $t=0$ to $t=1$ and from the $t=5$ to $t=7$. It moves backward from $t=1$ to $t=5$. The particle speeds up from $t=1$ to $t=2$ and from $t=5$ to $t=6$. It slows down from $t=0$ to $t=1$, from $t=3$ to $t=5$ and from $t=6$ to $t=7$. b) The acceleration is positive from $t=3$ to $t=6$. It is negative from $t=0$ to $t=2$ and from $t=6$ to $t=7$. It is zero from $t=2$ to $t=3$ and from $t=7$ to $t=9$. c) The particle moves at its greatest speed at $t=0$ and from $t=2$ to $t=3$. d) The particle stands still for more than an instant from $t=7$ to $t=9$.
a) The direction of the particle is decided by the sign of the velocity. In detail, - Positive velocity $(t=0$ to $t=1$ and $t=5$ to $t=7)$: the particle moves forward. - Negative velocity $(t=1$ to $t=5)$: the particle moves backward. Whether the particle speeds up or slows down depends on whether the velocity increases or decreases, but here the relationship is more complicated. In detail, *For positive velocities: - $t=5$ to $t=6$: the graph of the velocity goes up, meaning the particle speeds up. - $t=0$ to $t=1$ and $t=6$ to $t=7$: the graph of the velocity goes down, meaning the particle slows down. * For negative velocities: - $t=1$ to $t=2$: the graph of the velocity goes down, meaning the particle speeds up. - $t=3$ to $t=5$: the graph of the velocity goes up, meaning the particle slows down. b) Since acceleration is the slope of the tangent line to a point of the graph of velocity, whether it is positive or negative or zero lies in the direction of the graph of velocity at that point or that period. In detail, - $t=3$ to $t=6$: the graph of the velocity goes up, meaning the acceleration is positive. - $t=0$ to $t=2$ and $t=6$ to $t=7$: the graph of the velocity goes down, meaning the acceleration is negative. - $t=2$ to $t=3$ and $t=7$ to $t=9$: the graph of the velocity lies horizontally, meaning the acceleration there is zero. c) Since speed is the absolute value of velocity, the greatest speed can be the peak of the graph when $v\gt0$ or the bottom of the graph when $v\lt0$. Here, wee see that - When $v\gt0$: the peak is at $t=0$. - When $v\lt0$: the bottom is from $t=2$ to $t=3$. The absolute values of these peaks and bottoms look like they are equal, so we conclude that the particle moves at its greatest speed at $t=0$ and from $t=2$ to $t=3$. d) The particle stands still when its velocity is $0$. Here from $t=7$ to $t=9$, the velocity is $0$, so the particle stands till for $2$ seconds here.