Velocity is the rate of change of position with respect to time, whereas acceleration is the rate of change of velocity. Both are vector quantities (and so also have a specified direction), but the units of velocity are meters per second while the units of acceleration are meters per second squared.
A. Move 2 m east and then 12 m east; displacement is 14 m east and the distance is 14 m
B. Move 10 m east and then 12 m west, the displacement is 2 m west and the distance is 22 m.
C. Move 8 m west and then 16 m east; the displacement is 8 m east and the distance is 24 m
D. Move 12 m west and then 8 m east; the displacement is 4 m and the distance is 20 m
Answer:
ooh thanks but don't give me advise
(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.
Recall that
We have 
, so that
(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.
(c) The ball's average velocity is 0. Average velocity is given by 
, and we know that .
(d) The position of the ball 
at time 
is given by
Take the starting position to be the origin, 
. Then after 6 seconds,
so the ball is 42 m away from where it started.
We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:
Since the velocity is positive, the ball is still moving up the incline.
Answer:
73.5 m/s
Explanation:
The position of the first ball is:
y = y₀ + v₀ t + ½ at²
y = h + (0)(18) + ½ (-9.8)(18)²
y = h − 1587.6
The position of the second ball is:
y = y₀ + v₀ t + ½ at²
y = h + (-v) (18−6) + ½ (-9.8)(18−6)²
y = h − 12v − 705.6
Setting the positions equal:
h − 1587.6 = h − 12v − 705.6
-1587.6 = -12v − 705.6
1587.6 = 12v + 705.6
882 = 12v
v = 73.5
The second ball is thrown downwards with a speed of 73.5 m/s
