Question

different velocity-time graph lines are shown, displaying the behavior of multiple race carts along a linear track. which description correctly identifies cart c? a has a positive acceleration and is speeding up. b moving with zero acceleration. c moving at a constant acceleration. d has a negative acceleration and is slowing down.

Explanation:

To solve this, we analyze the velocity - time graph for cart C:

Step 1: Recall the relationship between velocity - time graph and acceleration

In a velocity - time graph, the slope of the line represents acceleration. The formula for acceleration \(a=\frac{\Delta v}{\Delta t}\), where \(\Delta v\) is the change in velocity and \(\Delta t\) is the change in time. If the slope is zero, acceleration \(a = 0\) (constant velocity, zero acceleration). If the slope is non - zero, there is non - zero acceleration.

Step 2: Analyze the slope of cart C's velocity - time line

Looking at the velocity - time graph for cart C, the line is vertical? No, wait, from the description, cart C's line: if the velocity - time graph line for cart C is vertical? No, actually, if the line is vertical, \(\Delta t = 0\) which is not possible. Wait, maybe the line for cart C is a vertical line? No, that can't be. Wait, maybe the line for cart C is a horizontal line? Wait, no, the options:

• Option A: Positive acceleration and speeding up. For positive acceleration, the slope of \(v - t\) graph should be positive (since \(a=\frac{\Delta v}{\Delta t}\), if \(\Delta v\) and \(\Delta t\) have the same sign, slope is positive). But if the cart is speeding up with positive acceleration, velocity and acceleration should have the same sign. But we need to check cart C's graph.
• Option B: Moving with zero acceleration. If the velocity - time graph is a horizontal line (constant velocity), then \(\Delta v=0\), so \(a = \frac{\Delta v}{\Delta t}=0\) (since \(\Delta t

eq0\)). So if cart C's velocity - time graph is a horizontal line, it has zero acceleration.

• Option C: Moving at a constant acceleration. A constant acceleration would be a straight line with non - zero slope. But if the line is horizontal (slope = 0), acceleration is zero, not constant non - zero.
• Option D: Negative acceleration and slowing down. For negative acceleration, slope is negative (\(\Delta v\) and \(\Delta t\) have opposite signs). But if the cart is slowing down, velocity and acceleration should have opposite signs. But we need to check cart C's graph.

From the velocity - time graph (even though the graph is described, the key is that for a velocity - time graph, a horizontal line means constant velocity, so acceleration \(a = 0\). So cart C has zero acceleration.

Answer:

B. Moving with zero acceleration.