Answer:
I. 0 m/s
II. 20 m/s
III. Part BC
Explanation:
I. Determination of the initial velocity.
From the diagram given above,
The motion of the car starts from the origin. This implies that the car start from rest and as such, the initial velocity of the car is 0 m/s
II. Determination of the maximum velocity attained.
From the diagram given above, we can see clearly that the maximum velocity is 20 m/s.
III. Determination of the part of the graph that represents zero acceleration.
It important that we know the meaning of zero acceleration.
Zero acceleration simply means the car is not accelerating. This can only be true when the car is moving with a constant velocity.
From the graph given above, the car has a constant velocity between B and C.
Therefore, part BC illustrates zero acceleration.
Answer:
t_{out} =
t_{in}, t_{out} = 
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is

The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D / 
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point

= D / 
D = v_{sg 2} t_{in}
with the distance is the same we can equalize

t_{out} = t_{in}
t_{out} =
t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D / 
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} = 
-- She went up for 0.4 sec and down for 0.4 sec.
-- The vertical distance traveled in gravity during ' t ' seconds is
D = (1/2) x (g) x (t)²
= (1/2) (9.8 m/s²) (0.4 sec)²
= (4.9 m/s²) x (0.16 s²)
= 0.784 meter ( B )
A table would be the most appropriate because that way you can compare the data.