In: Physics
This is a tricky question.a person doesnt actually change his
position (when compared to the horizontal distance he covered),
with respect to a fixed object(say a tree) when he does the squat .
His center of mass changes position. So the graph for position,
velocity and acceleration remains the same, if we take the
horizontal distance he covered.
but if we consider the change in position of his center of mass,
then things are different.
Since no specific numbers are given, lets assume that the person's center of mass is at about 1.5 metres above ground (this is going to be a very tall massive man!!). When in goes down into the squating position, lets assume that his center of mass is at about 0.5 metres above ground. then the position time graph for the above said exercise would look like the one given below (fig1)
(fig1)
The velocity time graph will be like fig2
(fig2)
As he goes down, his velocity increases rapidly and then reaches a
peak velocity and then decreases to zero. Since he is going down,
his velocity is considered as negative. as u can see from the
figure, in natural cases, the acceleration will be higher as we
begin to move down or as we begin to move up. this is why the graph
is asymemtric .
the velocity time graph will look like fig3
(fig3)
As he starts to move down, he starts to accelerate in the negative y direction, so the acceleration is negative. and after reaching the peak velocity, he then starts to decelerate and then comes to a zero velocity. When he moves up, the same process takes place, but in the opposite direction, and in much shorter time. So the upward motion will have a higher acceleration, and the peak will be higher compared to the value when he was coming down.
As you might've noticed, the y axis values are not given. this is because there are no real numbers given for actual calculation