For the given all options we can find the respective quantity
whenever we have to relate slope of a graph then it is always derivative of quantity on Y axis with respect to X axis
while if we need to find the area of the graph then it is integral of quantity on Y axis with respect to x axis
A. The area under that object's velocity-time graph
this shows area of velocity time graph = [tex]\int v dt[/tex] = displacement
so its not correct
B. The slope of that object's position-time graph
This shows slope of position time graph = [tex]\frac{ds}{dt}[/tex] = velocity
so this is correct for velocity
C. The slope of that object's velocity-time graph
This shows slope of velocity time graph = [tex]\frac{dv}{dt}[/tex] = acceleration
D. The area under that object's position-time graph
this shows the area of position time graph = [tex]\int s dt[/tex]
this is no physical quantity
So option B is correct
