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horsena [70]
4 years ago
14

Name the physical quantity measured by the area under velocity time graph?

Physics
1 answer:
Sophie [7]4 years ago
5 0
The area under a speed-time graph, between two time marks and between the
x-axis and the graph line, is exactly the distance traveled during that time.
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Two campers dock a canoe. One camper has a mass of 100.0 kg and moves forward at 3.0 m/s as he leaves the canoe to step onto the
KengaRu [80]

Answer:

V = -1.33 m/s

Explanation:

Given,

The mass of the camper, m = 100 kg

The velocity of the camper, v = 3.0 m/s

The combined mass of canoe and another camper, M = 225 kg

The velocity of the combined canoe and another camper, V = ?

According to the law of conservation of momentum,

                          MV + mv = 0  (Since the initial momentum of the dock is 0)

                             V = -mv/M

                                =  -100 x 3/225

                                = -1.33 m/s

The negative sign indicates that the combined objects moves opposite to that of the camper.

Hence, the velocity of the combined canoe and camper is, V = -1.33 m/s

6 0
4 years ago
Physics , help please
zaharov [31]
Acceleration in a velocity vs time graph is just the slope at that point. The reason for that is because the definition of acceleration is the change in velocity per unit of time. In this case we want instantaneous, which is the derivative or tangent line at that point.

At 3s we can see the slope is 0, so that means his acceleration is zero. That means he was moving at a constant velocity

At 5s we can see that the slope is negative. And from 5s to 6s the change in velocity is -5m/s^2

At 7s we can see the slope is very positive. And from 7s to 8s the change in velocity is +15m/s^2

And again, at 9s the slope is 0 so his acceleration is also zero. He’s moving at a constant velocity

If you take the integral of a velocity vs time graph, you get position. So the area underneath a velocity vs time graph is the distance traveled. Anything below the x axis is considered negative distance. We need to take the area of a triangle and the area of two rectangles to find the distance.

So, let’s do the two rectangles first. From 8s to 9s it is a width of 1 and a length of 40. So the area would be 40 meters. Let’s do the second rectangle. From 7s to 8s it is a width of 1. Then the length goes up to 25. So the area is 25 meters.

Now the triangle, the base is 1 and the height is 15. Divide 15 in half to get 7.5 meters

25 + 40 + 7.5 = 72.5 meters
6 0
3 years ago
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