using graph paper: for each graph, graph the missing two graphs. every question should have a x vs. t, v vs…

Explanation:

Step1: Recall kinematic - graph relationships

The slope of a position - time ($x - t$) graph gives velocity, and the slope of a velocity - time ($v - t$) graph gives acceleration. Displacement can be found from the area under the $v - t$ graph and average velocity is displacement divided by time interval. Instantaneous velocity is the slope of the $x - t$ graph at a point.

Question 1a

• $t = 0 - 3s$: The $x - t$ graph has a negative slope, so the object is moving in the negative direction. The magnitude of the slope is decreasing, so the object is slowing down.
• $t = 3 - 5s$: The slope of the $x - t$ graph is zero, so the object is at rest.
• $t = 5 - 8s$: The $x - t$ graph has a positive slope, so the object is moving in the positive direction. The magnitude of the slope is increasing, so the object is speeding up.

Question 1b

• $t = 0 - 3s$:
  • Displacement $\Delta x=x(3)-x(0)$. From the graph, if we assume $x(0) = 0$ and $x(3)\approx - 3$, then $\Delta x=-3 - 0=-3$.
  • Average velocity $v_{avg}=\frac{\Delta x}{\Delta t}=\frac{-3}{3}=- 1$.
• $t = 3 - 5s$:
  • Displacement $\Delta x=x(5)-x(3)$. Since the object is at rest, $\Delta x = 0$.
  • Average velocity $v_{avg}=\frac{\Delta x}{\Delta t}=0$.
• $t = 5 - 8s$:
  • Let $x(5)\approx - 3$ and $x(8)\approx3$. Then $\Delta x=x(8)-x(5)=3-( - 3)=6$.
  • Average velocity $v_{avg}=\frac{\Delta x}{\Delta t}=\frac{6}{3}=2$.

Question 1c

• $t = 1s$: The slope of the $x - t$ graph at $t = 1s$ is approximately $\frac{-1 - 0}{1}=-1$. So the instantaneous velocity $v(1)=-1$.
• $t = 4s$: Since the object is at rest from $t = 3 - 5s$, $v(4)=0$.
• $t = 6s$: The slope of the $x - t$ graph at $t = 6s$ is positive. If we consider two - point on the graph around $t = 6s$, say $(5,-3)$ and $(7,0)$, the slope is $\frac{0 + 3}{7 - 5}=1.5$. So $v(6)=1.5$.

Question 1d

The total displacement between $t = 0$ and $t = 8s$ is $x(8)-x(0)$. If $x(0) = 0$ and $x(8)\approx3$, then $\Delta x=3$.

Question 2a

• $t = 0 - 2s$: The $v - t$ graph is positive and sloping downwards, so the object is moving in the positive direction and slowing down.
• $t = 2 - 4s$: The $v - t$ graph is negative and sloping downwards, so the object is moving in the negative direction and speeding up.
• $t = 4 - 6s$: The $v - t$ graph is negative and constant, so the object is moving in the negative direction with a constant speed.
• $t = 6 - 7s$: The $v - t$ graph is negative and sloping upwards, so the object is moving in the negative direction and slowing down.
• $t = 7 - 8s$: The $v - t$ graph is positive and sloping upwards, so the object is moving in the positive direction and speeding up.

Question 2b

• $t = 0 - 2s$:
  • Displacement is the area under the $v - t$ graph. The area of the trapezoid is $\Delta x=\frac{(1 + 2)\times2}{2}=3$.
  • Average acceleration $a_{avg}=\frac{v(2)-v(0)}{2}=\frac{0 - 2}{2}=-1$.
• $t = 2 - 4s$:
  • The area of the triangle is $\Delta x=\frac{(0 + 2)\times2}{2}= - 2$.
  • Average acceleration $a_{avg}=\frac{v(4)-v(2)}{2}=\frac{-2-0}{2}=-1$.
• $t = 4 - 6s$:
  • Displacement $\Delta x=-2\times2=-4$.
  • Average acceleration $a_{avg}=0$ (since $v$ is constant).
• $t = 6 - 7s$:
  • Displacement $\Delta x=\frac{(-2+0)\times1}{2}=-1$.
  • Average acceleration $a_{avg}=\frac{v(7)-v(6)}{1}=\frac{0 + 2}{1}=2$.
• $t = 7 - 8s$:
  • Displacement $\Delta x=\frac{(0 + 2)\times1}{2}=1$.
  • Average acceleration $a_{avg}=\frac{v(8)-v(7)}{1}=\frac{2-0}{1}=2$.

Question 2c

• $t = 1s$: The slope of the $v - t$ graph at $t = 1s$ is $\frac{0 - 2}{2}=-1$. So $a(1)=-1$.
• $t = 5s$: Since the $v - t$ graph is flat, $a(5)=0$.
• $t = 7s$: The slope of the $v - t$ graph at $t = 7s$ is $\frac{2-0}{1}=2$. So $a(7)=2$.

Question 2d

The total displacement is the sum of the displacements in each interval: $\Delta x=3-2 - 4-1 + 1=-3$.

Question 3a

We know that $v=v_0+\int_{0}^{t}a\mathrm{d}t$. Given $v_0 = 4$, and the area under the $a - t$ graph from $t = 0$ to $t = 3$ is $\int_{0}^{3}a\mathrm{d}t=\frac{(1 + 0)\times3}{2}=1.5$. So $v(3)=v_0+\int_{0}^{3}a\mathrm{d}t=4 + 1.5=5.5$.

Question 3b

The object's velocity is given by $v = v_0+\int_{0}^{t}a\mathrm{d}t$. The area under the $a - t$ graph from $t = 0$ to any time $t$ in the given interval is always non - negative when added to the initial positive velocity $v_0 = 4$. So the object is never moving in the negative direction.

Answer:

Question 1a

• $t = 0 - 3s$: Moving in negative direction, slowing down.
• $t = 3 - 5s$: At rest.
• $t = 5 - 8s$: Moving in positive direction, speeding up.

Question 1b

• $t = 0 - 3s$: Displacement $=-3$, Average velocity $=-1$.
• $t = 3 - 5s$: Displacement $=0$, Average velocity $=0$.
• $t = 5 - 8s$: Displacement $=6$, Average velocity $=2$.

Question 1c

• $t = 1s$: $v=-1$
• $t = 4s$: $v = 0$
• $t = 6s$: $v = 1.5$

Question 1d

$3$

Question 2a

• $t = 0 - 2s$: Moving in positive direction, slowing down.
• $t = 2 - 4s$: Moving in negative direction, speeding up.
• $t = 4 - 6s$: Moving in negative direction, constant speed.
• $t = 6 - 7s$: Moving in negative direction, slowing down.
• $t = 7 - 8s$: Moving in positive direction, speeding up.

Question 2b

• $t = 0 - 2s$: Displacement $=3$, Average acceleration $=-1$.
• $t = 2 - 4s$: Displacement $=-2$, Average acceleration $=-1$.
• $t = 4 - 6s$: Displacement $=-4$, Average acceleration $=0$.
• $t = 6 - 7s$: Displacement $=-1$, Average acceleration $=2$.
• $t = 7 - 8s$: Displacement $=1$, Average acceleration $=2$.

Question 2c

• $t = 1s$: $a=-1$
• $t = 5s$: $a = 0$
• $t = 7s$: $a = 2$

Question 2d

$-3$

Question 3a

$5.5$

Question 3b

No. The area under the $a - t$ graph added to the initial positive velocity always gives a non - negative velocity.