Question

question 4 1 pts three carts move along the x-axis. the locations and instantaneous velocities of the carts are shown in the figure. which of the following correctly ranks the carts according to the magnitude of their position vectors relative to the origin at the instant shown?

○ b>a>c

○ c>b>a

○ c>a>b

○ b>c>a

Explanation:

Step1: Determine position of Cart A

Cart A is at \( x = 0 \), so the magnitude of its position vector \( |\vec{r}_A|=|0| = 0\space cm \).

Step2: Determine position of Cart B

From the figure, Cart B is between \( x = 0 \) and \( x = 40 \space cm \), let's assume each tick is \( 10 \space cm \) (since from \( - 40 \) to \( 0 \) and \( 0 \) to \( 40 \) there are 4 ticks each? Wait, no, looking at the first graph, Cart A is at \( x = 0 \), the second graph: Cart B is at, let's see, the distance from \( 0 \) to B: if each interval is \( 10 \space cm \) (since \( - 40 \) to \( 0 \) has 4 intervals? Wait, no, \( x=-40 \), then next tick, then next, then next, then \( x = 0 \), so each interval is \( 10 \space cm \) (because \( 40\space cm/4 = 10\space cm \) per interval). So Cart B: from \( x = 0 \), moving 2 intervals to the right? Wait, no, the second graph: Cart B is at a position where from \( x = 0 \), the first tick is \( 10 \), second \( 20 \)? Wait, maybe better to estimate: Cart A is at \( x = 0 \), Cart B is at, say, \( x = 20 \space cm \) (since between \( 0 \) and \( 40 \), two ticks? Wait, the first graph: \( x=-40 \), then three ticks, then \( x = 0 \), then three ticks to \( x = 40 \). So each tick is \( 10 \space cm \) (because \( 40/4 = 10 \)? Wait, \( - 40 \) to \( 0 \): 4 intervals (from \( - 40 \) to \( - 30 \), \( - 30 \) to \( - 20 \), \( - 20 \) to \( - 10 \), \( - 10 \) to \( 0 \)), so each interval is \( 10 \space cm \). Then Cart B: in the second graph, Cart B is at \( x = 20 \space cm \) (two intervals from \( 0 \) to \( 20 \)). Cart C: in the third graph, Cart C is at \( x=-30 \space cm \) (one interval left of \( - 40 \)? Wait, no: \( x=-40 \), then next tick is \( - 30 \), then Cart C is at \( - 30 \space cm \)? Wait, no, the third graph: Cart C is at a position where it's one interval to the right of \( x=-40 \), so \( x=-30 \space cm \)? Wait, no, the third graph: the cart C is at \( x=-30 \space cm \) (since \( x=-40 \), then a tick, then cart C). Wait, maybe I made a mistake. Let's re - examine:

Cart A: \( x = 0 \), so \( |x_A|=0 \)

Cart B: Let's see the second graph. The origin \( x = 0 \), then the first tick to the right is \( 10 \), second \( 20 \), third \( 30 \), fourth \( 40 \). Cart B is at the second tick? Wait, no, the arrow for B is at a position that is 2 intervals from \( 0 \), so \( x = 20 \space cm \), so \( |x_B| = 20 \space cm \)

Cart C: Third graph: \( x=-40 \), then a tick (so \( x=-30 \)), then cart C. So \( x_C=-30 \space cm \), so \( |x_C| = 30 \space cm \)

Wait, but maybe the intervals are different. Wait, another approach: the position vector's magnitude is the absolute value of the x - coordinate.

Cart A: at \( x = 0 \), so \( |\vec{r}_A|=0 \)

Cart B: Let's look at the second diagram. The cart B is at a position where the distance from the origin is less than \( 40 \), more than \( 0 \). Let's assume that each small division is \( 10 \space cm \). So from \( 0 \) to \( 40 \), there are 4 divisions (since \( 40/4 = 10 \)). So Cart B is at \( x = 20 \space cm \) (2 divisions from 0), so \( |x_B| = 20 \)

Cart C: In the third diagram, Cart C is at \( x=-30 \space cm \) (1 division to the right of \( - 40 \)), so \( |x_C| = 30 \)

Wait, but maybe I messed up Cart C's position. Wait, the third diagram: \( x=-40 \), then a tick, then cart C. So the coordinate of C is \( x=-30 \space cm \) (because \( - 40+10=-30 \)). Cart B: in the second diagram, cart B is at \( x = 20 \space cm \) ( \( 0 + 2\times10 = 20 \) ). Cart A: \( x = 0 \)

So magnitude of position vectors: \( |x_C|=30 \), \( |x_B| = 20 \), \( |x_A|=0 \)

Wait, but the options are:

B > A > C? No, because \( |x_C| = 30 \), \( |x_B| = 20 \), \( |x_A| = 0 \), so \( C > B > A \)? Wait, no, wait maybe my estimation of the positions is wrong.

Wait, let's look again:

First graph (Cart A): at \( x = 0 \), so \( x_A = 0 \), \( |x_A| = 0 \)

Second graph (Cart B): The cart is at a position that is 2 units to the right of \( 0 \), if each unit is \( 10 \space cm \), then \( x_B=20 \space cm \), \( |x_B| = 20 \)

Third graph (Cart C): The cart is at a position that is 1 unit to the right of \( - 40 \), so \( x_C=-30 \space cm \), \( |x_C| = 30 \)

Wait, but the options include C > B > A? Let's check the options:

Option 1: B > A > C → \( 20>0 > 30 \)? No

Option 2: C > B > A → \( 30>20 > 0 \)? Yes

Option 3: C > A > B → \( 30>0 > 20 \)? No

Option 4: B > C > A → \( 20>30 > 0 \)? No

Wait, but maybe I got Cart B's position wrong. Let's re - examine the second graph. The cart B is at a position where the distance from \( 0 \) is, say, \( 10 \space cm \)? No, the arrow for B is at a position that is between \( 0 \) and \( 40 \), closer to \( 20 \)? Wait, maybe the intervals are \( 10 \space cm \) per division. So:

Cart A: \( x = 0 \), \( |x| = 0 \)

Cart B: Let's count the divisions from \( 0 \) to B. In the second graph, from \( x = 0 \), the first division is \( 10 \), second \( 20 \), third \( 30 \), fourth \( 40 \). Cart B is at the second division? So \( x = 20 \), \( |x| = 20 \)

Cart C: In the third graph, from \( x=-40 \), the first division to the right is \( - 30 \), so Cart C is at \( x=-30 \), \( |x| = 30 \)

So the magnitudes are \( |x_C| = 30 \), \( |x_B| = 20 \), \( |x_A| = 0 \), so the order from largest to smallest is \( C > B > A \)

Answer:

C>B>A