Consider this situation: A car moves with a velocity of 2m/s for 10 seconds, and then its veocity increases to 10m/s in next 5 seconds. Thereafter its velocity decreases at a uniform rate until it comes to rest after 10 seconds a. Express this this entire run of car on the velocity time graph. b. From the graph identify the- 1)time interval when the car was accelerating 2)displacement traveed while the car was decelerating
Asked by Ayushi Joshi | 31st May, 2013, 06:46: PM
1. Velocity (y -axis) - time graph(x-axis) would be a line parallel to x axis at y = 2 m/s from t = 0 to t = 10 sec and then, it would be a straight line with positive slope between t = 10, y = 2m/s and t = 15 sec, y = 10m/s and then it will be a straight line with negative slope between t = 15, y = 10m/s and t = 25 sec, y = 0m/s.
2. So, the car was accelerating between 10 and 15 sec. A positive slope on the v-t graph shows acceleration.
3. Displacement = Area under the v-t graph.
To find the displacement when the car was decelerating, we need to find the area under the graph for the negative slope between t= 15 sec and t = 25 sec.
Area = Area of triangle so formed = 1/2*base*height = 1/2*(10sec) * 10ms = 50 m.
Answered by | 31st May, 2013, 07:44: PM
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