Answer:
See the answers below.
Explanation:
To solve this problem we must use the following equation of kinematics.

where:
y - yo = Fall distance [m]
Vo = initial velocity = 0 (dropped)
t = time = 6.5 [s]
g = gravity acceleration = 9.8 [m/s²]
Now replacing:
![y-y_{o}=0 +\frac{1}{2} *9.8*6.5^{2}\\y-y_{o}=207 [m]](https://tex.z-dn.net/?f=y-y_%7Bo%7D%3D0%20%2B%5Cfrac%7B1%7D%7B2%7D%20%2A9.8%2A6.5%5E%7B2%7D%5C%5Cy-y_%7Bo%7D%3D207%20%5Bm%5D)
The question is related to the distance between the points where the ball was dropped and the ground.
The variables used are gravitational acceleration, time, fall distance, initial velocity.
Answer:
Between 0 and 1 seconds (B)
Explanation:
The velocity of the car over time is represented by the line graphed here
the steeper the line, the greater change in velocity that occurred in a given time frame.
The steepest portion of the line is between 0-1 seconds, which means that the greatest rate of change occurred between 0-1 seconds.
(acceleration is the rate of change)