Answer:
<em>The distance is 35 m and the magnitude of the displacement is 26.93 m</em>
Explanation:
<u>Displacement and Distance</u>
These are two related concepts. A moving object constantly travels for some distance at defined periods of time. The total distance is the sum of each individual distance the object traveled. It can be written as:
dtotal=d1+d2+d3+...+dn
This sum is calculated independently of the direction the object moves.
The displacement only takes into consideration the initial and final positions of the object. The displacement, unlike distance, is a vectorial magnitude and can even have magnitude zero if the object starts and ends the movement at the same point.
Taylor walks 25 m north and 10 m west. The total distance is the sum of both numbers:
d = 25 m + 10 m = 35 m
To calculate the displacement, we need to know the final position with respect to the initial position. If we set the coordinates of Taylor's car as the origin (0,0), then his final position is (-10,25), assuming the west direction is negative and the north direction is positive.
The magnitude of the displacement is the distance from (0,0) to (-10,25):


D = 26.93 m
The distance is 35 m and the magnitude of the displacement is 26.93 m
It depends on what type of solid
I believe that the answer to this would be B
Hope this helped
Answer:
A. speed = 7.14 Km/s
B. distance = 1820.7 Km
Explanation:
Given that: a = 14.0 m/
, t = 8.50 minutes.
But,
t = 8.50 = 8.50 x 60
= 510 seconds
A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;
v = u + at
where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.
u = 0
So that,
v = 14 x 510
= 7140 m/s
The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.
B. the distance traveled can be determined by applying second equation of motion.
s = ut +
a
where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.
u = 0
s =
a
=
x 14 x 
= 7 x 260100
= 1820700 m
The distance that the shuttle has traveled during the given time is 1820.7 Km.