Question

b) calculate the slope of each of the x vs t graphs.

1. each one of the graphs below shows how the position of an object changes with time.

a
calculate the speed:
speed in m/s
position at t = 15s is:
position at t = 25s is:
b
calculate the speed:
speed in m/s
position at t = 10 min is:

Explanation:

Step1: Recall speed - slope relation

The speed of an object in an x - t graph is equal to the slope of the graph. The formula for the slope $m$ of a line is $m=\frac{\Delta x}{\Delta t}$, where $\Delta x$ is the change in position and $\Delta t$ is the change in time.

For graph A:

Step2: Select two points on graph A

Let's take two points on the line of graph A: $(t_1,x_1)=(0,0)$ and $(t_2,x_2)=(30,100)$.

Step3: Calculate the slope (speed) for graph A

Using the slope formula $m=\frac{\Delta x}{\Delta t}=\frac{x_2 - x_1}{t_2 - t_1}=\frac{100 - 0}{30 - 0}=\frac{10}{3}\text{ m/s}\approx3.33\text{ m/s}$.

Step4: Calculate position at $t = 15\text{ s}$ for graph A

Since the equation of the line is $x=vt$ (because $x_0 = 0$), and $v=\frac{10}{3}\text{ m/s}$, when $t = 15\text{ s}$, $x=v\times t=\frac{10}{3}\times15 = 50\text{ m}$.

Step5: Calculate position at $t = 25\text{ s}$ for graph A

$x=v\times t=\frac{10}{3}\times25=\frac{250}{3}\approx83.33\text{ m}$.

For graph B:

Step6: Select two points on graph B

Let's take two points on the line of graph B: $(t_1,x_1)=(0,0)$ and $(t_2,x_2)=(10,300)$.

Step7: Calculate the slope (speed) for graph B

Using the slope formula $m=\frac{\Delta x}{\Delta t}=\frac{x_2 - x_1}{t_2 - t_1}=\frac{300 - 0}{10 - 0}=30\text{ m/min}$. To convert to m/s, we know that $1\text{ min}=60\text{ s}$, so $v = 30\div60 = 0.5\text{ m/s}$.

Step8: Calculate position at $t = 10\text{ min}$ for graph B

From the graph, when $t = 10\text{ min}$, $x = 300\text{ m}$.

Answer:

For graph A:

Speed: $\frac{10}{3}\text{ m/s}\approx3.33\text{ m/s}$
Position at $t = 15\text{ s}$: $50\text{ m}$
Position at $t = 25\text{ s}$: $\frac{250}{3}\text{ m}\approx83.33\text{ m}$

For graph B:

Speed: $0.5\text{ m/s}$
Position at $t = 10\text{ min}$: $300\text{ m}$