How Do You Calculate Position from a Velocity Graph?

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To calculate the position from a velocity graph, the car's initial position is 10m at t = 0s. At t = 2s, the velocity is determined to be 4m/s, leading to a position calculation of 18m using the formula x = vt + x0. The discussion highlights that the car is decelerating, indicated by a negative slope on the velocity graph, suggesting it changes direction when the velocity becomes negative. Integration of the velocity function is recommended for a more accurate position calculation over time. Understanding these concepts is crucial for solving similar physics problems effectively.
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Homework Statement



5. A car starts from xi = 10m at ti = 0s and moves with the velocity graph shown in figure on the right.
a. What is the object’s position at t = 2s, 3s, and 4s?
b. Does this car ever change direction? If so, at what time?

Homework Equations



V = d/t

The Attempt at a Solution



for t at 2 secs...
according to graph v= 4m/s

so position (d) = (t)(v)
= (2)(4)
= 8m?

am i doing it right?


OR

v = change d/change t
4m/s = (x-10)/(2-0)
4m/s = (x-10)/(2)
8m/s = x-10
x = 18m

is that correct?
 

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for t at 2 secs...
according to graph v= 4m/s

so position (d) = (t)(v)
= (2)(4)
= 8m?

am i doing it right?
Well this can't be right if the car starts at 10 m and adds distance. The slope of the plot of velocity vs time is negative, which indicates the car is decelerating, and the since the slope is constant, the deceleration is constant.

Certainly when the car has a negative velocity, it is reversing.

Is one familiar with integration?

if v(t) = d x(t)/dt, then

x(t) = \int_0^t\,v(t) dt\,+\,x(0)

This might be useful:

http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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