Question

the graph below shows vₓ versus t for an object moving along a straight line. how would you find the displacement δx of the object for the first 5.0 s of the motion? multiple choice
use the average slope of the graph between t = 0.0 s to t = 14 s.
compute the area under the graph from t = 0.0 s to t = 14 s.
compute the area under the graph from t = 0.0 s to t = 5.0 s.
use the slope of the graph between t = 0.0 s to t = 5.0 s.

Explanation:

In a velocity - time (\( v - t \)) graph, the displacement \( \Delta x \) of an object is equal to the area under the \( v - t \) graph between the corresponding time intervals. To find the displacement for the first \( 5.0 \, \text{s} \) of motion, we need to calculate the area under the \( v - t \) graph from \( t = 0.0 \, \text{s} \) to \( t = 5.0 \, \text{s} \). The slope of a \( v - t \) graph gives acceleration, not displacement, so calculating the slope is incorrect. And calculating the area from \( t = 0.0 \, \text{s} \) to \( t = 14 \, \text{s} \) would give the displacement for the entire \( 14 \, \text{s} \) interval, not just the first \( 5.0 \, \text{s} \).

Answer:

Compute the area under the graph from \( t = 0.0 \, \text{s} \) to \( t = 5.0 \, \text{s} \)