Question

1. the graph below shows the velocity v = f(t) of a particle moving on a coordinate line.

a) when does the particle move forward? when does it move backward?
b) when is the particle speeding up? when is it slowing down?
c) when does the particle move at its greatest speed?
d) when does the particle move at a constant speed?
e) when does the particle stand still for more than an instant?
f) describe what the graph of the body’s speed would look like?
g) when is the particle’s acceleration positive? when is it negative? when is it zero? what would this mean for the graph of the body’s acceleration if you were to graph it?

Explanation:

Step1: Recall velocity - direction relation

A particle moves forward when velocity \(v>0\) and backward when \(v < 0\).
From the graph, \(v>0\) on the intervals \((0,2)\cup(5,8)\) and \(v < 0\) on the intervals \((2,5)\).

Step2: Recall velocity - acceleration relation for speeding up/slowing down

A particle is speeding up when \(v\) and \(a\) (acceleration, slope of \(v - t\) graph) have the same sign, and slowing down when they have opposite signs.
On \((0,1)\cup(5,6)\), \(v>0\) and \(a>0\) (positive - slope), on \((3,4)\), \(v < 0\) and \(a<0\) (negative - slope), so it's speeding up on \((0,1)\cup(3,4)\cup(5,6)\). On \((1,2)\cup(4,5)\cup(6,8)\), \(v\) and \(a\) have opposite signs, so it's slowing down on \((1,2)\cup(4,5)\cup(6,8)\).

Step3: Find maximum speed

Speed is the absolute - value of velocity. The greatest value of \(|v|\) occurs at \(t = 1\) and \(t = 6\) (highest peaks of the \(v - t\) graph).

Step4: Identify constant speed

Constant speed means \(|v|\) is constant. This occurs on \((3,4)\) and \((7,8)\) where the velocity is constant (horizontal line segments).

Step5: Find when particle is still

The particle is still when \(v = 0\). This occurs at \(t=2\), \(t = 5\), and \(t = 8\), but for more than an instant at \(t\in[3,4]\) and \(t\in[7,8]\).

Step6: Describe speed graph

The speed graph \(s(t)=|v(t)|\). It will be the same as the \(v - t\) graph for \(v\geq0\) and the reflection of the \(v - t\) graph about the \(t\) - axis for \(v < 0\). So it will have non - negative values only.

Step7: Analyze acceleration

Acceleration \(a\) is the slope of the \(v - t\) graph. \(a>0\) on \((0,1)\cup(5,6)\), \(a < 0\) on \((1,2)\cup(6,8)\), and \(a = 0\) on \((3,4)\cup(7,8)\) (horizontal segments). The acceleration graph will have positive values on \((0,1)\cup(5,6)\), negative values on \((1,2)\cup(6,8)\), and zero values on \((3,4)\cup(7,8)\).

Answer:

a) Moves forward on \((0,2)\cup(5,8)\), moves backward on \((2,5)\).
b) Speeding up on \((0,1)\cup(3,4)\cup(5,6)\), slowing down on \((1,2)\cup(4,5)\cup(6,8)\).
c) At \(t = 1\) and \(t = 6\).
d) On \((3,4)\) and \((7,8)\).
e) On \([3,4]\) and \([7,8]\).
f) Non - negative graph, same as \(v - t\) for \(v\geq0\) and reflection of \(v - t\) about \(t\) - axis for \(v < 0\).
g) \(a>0\) on \((0,1)\cup(5,6)\), \(a < 0\) on \((1,2)\cup(6,8)\), \(a = 0\) on \((3,4)\cup(7,8)\). The acceleration graph will have positive segments on \((0,1)\cup(5,6)\), negative segments on \((1,2)\cup(6,8)\) and zero segments on \((3,4)\cup(7,8)\).