Question

question 1 (1 point)
the slope of a position versus time graph gives
position.
velocity.
acceleration.
displacement.
question 2 (1 point)
an object moves 15.0 m north and then 11.0 m south. find both the distance traveled and the magnitude of the displacement vector.
4.0 m, 26.0 m
26.0 m, 4.0 m
26.0 m, 26.0 m
4.0 m, 4.0 m
question 3 (1 point)

Explanation:

Question 1

In kinematics, the slope of a position - time graph is calculated as the change in position ($\Delta x$) divided by the change in time ($\Delta t$), and velocity ($v$) is defined as $v=\frac{\Delta x}{\Delta t}$. Position is the y - axis variable, acceleration is the slope of a velocity - time graph, and displacement is a change in position. So the slope of a position - versus - time graph gives velocity.

Step 1: Calculate the distance traveled

Distance is the total length of the path taken. The object moves 15.0 m north and then 11.0 m south. So we add these two distances: $d = 15.0\ m+ 11.0\ m=26.0\ m$.

Step 2: Calculate the magnitude of displacement

Displacement is the straight - line distance from the initial position to the final position. Let's take north as positive. The displacement $\Delta x=x_2 - x_1$. The object moves 15.0 m north (so $x_1 = 0$, $x_2$ after first move is 15.0 m) and then 11.0 m south (so the final position $x_3=15.0\ m - 11.0\ m = 4.0\ m$ from the initial position). The magnitude of displacement is 4.0 m.

Answer:

B. velocity

Question 2