It then continues in the same direction for another 40 km at 60 km/h. The second graph also consists of two (solid) line segments, having the same slopes as before - the main difference (compared to the first graph) being that the stage involving higher-speed motion lasts much longer.Īn automobile travels on a straight road for 40 km at 30 km/h. The slope of the dashed line represents the average velocity (in both graphs). The first consists of two (solid) line segments, the first having a slope of 1.22 and the second having a slope of 3.05. (c) The graphs (Fig.1a) and (Fig.1b) are shown below (with meters and seconds understood). (b) Using the fact that distance = vt while the velocity v is constant, we find (a) Using the fact that time = distance/velocity while the velocity is constant, we find

(c) Graph x versus t for both cases and indicate how the average velocity is found on the graph. (b) You walk for 1.00 min at a speed of 1.22 m/s and then run for 1.00 min at 3.05 m/s along a straight track. Thus, in 0.50 s, the car travels a distance isĬompute your average velocity in the following two cases: (a) You walk 73.2 m at a speed of 1.22 m/s and then run 73.2 m at a speed of 3.05 m/s along a straight track. If you are driving a car at 90 km/h during such a sneeze, how far does the car move during that time? Problem#1During a hard sneeze, your eyes might shut for 0.50 s.

Positive physics graph problems