Question

velocity-time graph
what is the displacement in meters of the moving object from 0-20.0 s?
(enter just the numerical answer, with a sign if needed, but without the unit)
respuesta

Explanation:

Step1: Identify the shape (trapezoid)

From 0 - 10 s: triangle, 10 - 20 s: rectangle. Or combined as trapezoid with bases \( b_1 = 10 \, \text{s}, b_2 = 10 \, \text{s} \) (wait, no: time from 0 - 20 s. The velocity - time graph from 0 - 10 s is a triangle (wait, no, from 0 to 10 s, velocity goes from 0 to 10 m/s? Wait, the graph: from t=0, v=0; at t=10, v=10; then constant until t=20. So the area from 0 - 20 s is the area of the trapezoid (or triangle + rectangle). The formula for the area of a trapezoid is \( A=\frac{(a + b)}{2}\times h \), where \( a \) and \( b \) are the two parallel sides (time intervals) and \( h \) is the height (velocity). Wait, actually, in velocity - time graph, displacement is the area under the graph. From 0 - 10 s: it's a triangle with base \( t_1 = 10 \, \text{s} \) and height \( v = 10 \, \text{m/s} \). Area of triangle: \( \frac{1}{2}\times10\times10 = 50 \). From 10 - 20 s: it's a rectangle with length \( t_2 = 10 \, \text{s} \) and height \( v = 10 \, \text{m/s} \). Area of rectangle: \( 10\times10 = 100 \). Total area (displacement) = 50+100 = 150.
Alternatively, using trapezoid formula: the two parallel sides (in terms of time) are the time at 0 (0 s) and time at 20 s (20 s), but the velocity at 0 is 0, at 10 is 10, at 20 is 10. Wait, the graph from 0 - 10 is a line from (0,0) to (10,10), then horizontal to (20,10). So the shape is a trapezoid with bases \( b_1 = 10 \, \text{s} \) (the horizontal part from 10 - 20) and \( b_2 = 20 \, \text{s} \) (total time), but no, better to split into triangle and rectangle.
Wait, the velocity at t=0 is 0, t=10 is 10, t=20 is 10. So the area under the graph from 0 - 20 is the area of the triangle (0 - 10 s) plus the area of the rectangle (10 - 20 s).
Triangle area: \( \frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times10\times10 = 50 \)
Rectangle area: \( \text{length}\times \text{height}=10\times10 = 100 \)
Total displacement (area) = 50 + 100 = 150.

Step2: Sum the areas

Add the area of the triangle (0 - 10 s) and the rectangle (10 - 20 s).
\( 50+100 = 150 \)

Answer:

150