Question

1. what is the velocity of the rolling ball shown in this position - time graph?

a. 3 m/s w
b. 8 m/s w
c. 2 m/s w
d. 5 m/s w
e. 6 m/s w
short response (6 marks) - show all your work including the formula applied, remember to include units and directions where applicable. check your significant figures in your answers.

1. a plane is landing on a runway with an unknown initial velocity. the length of the runway is 3750 m e. if the plane comes to a complete stop 25.0 seconds after landing, what is the acceleration of the plane during this? 3 marks - t
2. determine the max height of a ball if it is thrown at 15 m/s up. 3 marks - t

sig figs & units check: 0.5 mark will be deducted for sig fig errors and missing units in final answers, up to 2 marks. 2 marks - c

Explanation:

Question 10

Step1: Recall velocity - position - time relation

Velocity $v$ in a position - time graph is the slope of the graph. The slope formula is $m=\frac{\Delta y}{\Delta x}$, where for a position - time graph, $\Delta y$ is the change in position and $\Delta x$ is the change in time.
Let's assume two points on the line: $(t_1,x_1)=(0,0)$ and $(t_2,x_2)=(4,8)$. Then $v = \frac{x_2 - x_1}{t_2 - t_1}=\frac{8 - 0}{4 - 0}=2$ m/s.

Step1: Identify the kinematic equation

We use the kinematic equation $x=v_0t+\frac{1}{2}at^{2}$ and $v = v_0+at$. Since the plane stops, $v = 0$. We can also use the equation $x=\frac{v_0 + v}{2}t$ to find $v_0$ first. Given $x = 375$ m and $t=25.0$ s and $v = 0$. From $x=\frac{v_0 + v}{2}t$, we have $375=\frac{v_0+0}{2}\times25$, so $v_0=\frac{2\times375}{25}=30$ m/s.

Step2: Use the acceleration formula

Then use $a=\frac{v - v_0}{t}$. Substitute $v = 0$, $v_0 = 30$ m/s and $t = 25.0$ s. So $a=\frac{0 - 30}{25}=- 1.2$ m/s².

Step1: Identify the kinematic equation

At the maximum - height, the final velocity $v = 0$. We use the kinematic equation $v^{2}-v_0^{2}=2ah$. Given $v_0 = 15$ m/s, $v = 0$ and $a=-g=- 9.8$ m/s².

Step2: Solve for height $h$

From $v^{2}-v_0^{2}=2ah$, we can re - arrange it to $h=\frac{v^{2}-v_0^{2}}{2a}$. Substitute the values: $h=\frac{0-(15)^{2}}{2\times(-9.8)}=\frac{- 225}{-19.6}\approx11.5$ m.

Answer:

c. 2 m/s [W]

Question 11