A body moving with uniform acceleration has two points (15,15) and (20,16) on the velocity -time graph of its motion. Calculate the distance between the two points?
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assuming (x,y) y is time, x is distance.
V =(20-15)/(16-15)=5m/s
First, can you calculate the acceleration given the two data points?
To calculate the distance between the two points on the velocity-time graph, we need to use the equation for average acceleration:
average acceleration = change in velocity / change in time
Given two points (15,15) and (20,16) on the velocity-time graph, we can calculate the change in velocity and change in time between these two points.
Change in velocity = final velocity - initial velocity
Initial velocity is the velocity at the first point, which is 15.
Final velocity is the velocity at the second point, which is 16.
So, the change in velocity is 16 - 15 = 1 m/s.
Change in time = final time - initial time
In this case, the time scale is not given, so we can use the difference in x-coordinates between the two points as the change in time.
Initial time corresponds to the x-coordinate of the first point, which is 15.
Final time corresponds to the x-coordinate of the second point, which is 20.
So, the change in time is 20 - 15 = 5 seconds.
Now, we have the change in velocity and change in time, so we can calculate the average acceleration:
average acceleration = change in velocity / change in time
average acceleration = 1 m/s / 5 seconds
average acceleration = 0.2 m/s²
The distance covered by a body under uniform acceleration can be calculated using the equation:
distance = (initial velocity * time) + (1/2 * acceleration * time^2)
Given that the initial velocity is 15 m/s and the average acceleration is 0.2 m/s², and the change in time between the two points is 5 seconds, we can substitute these values into the equation:
distance = (15 * 5) + (1/2 * 0.2 * 5^2)
distance = 75 + (1/2 * 0.2 * 25)
distance = 75 + 0.5 * 0.2 * 25
distance = 75 + 0.1 * 25
distance = 75 + 2.5
distance = 77.5 meters
Therefore, the distance between the two points on the velocity-time graph is 77.5 meters.