10 Km.
S= Speed
D= distance
T= time
S= d/t
but since you are solving for "d" the equation is d=st so you plug in 10 km/h for speed and 2.1 hours for time and just multiply them. The hours cancel out so you are left with 10km.
Answer:
d = 13 miles
Explanation:
Lets say the position of court house is origin in this case
her office is located at 4 miles west and 4 miles south of court house
so here we have coordinate of the office with respect to court house is given as

now the position of her home is located at 1 miles east and 8 miles north of the court house
so the coordinates of her home is given as

now the change in the position is given as the distance between office and home



Answer:
The answer depends on what object you are dropping. Are you dropping a balloon or a car? (I'm joking 'bout that one.) If the mass of the object is very little, then it might drop slower. If the mass is bigger, then it might drop faster.
Good luck!
Explanation:
<u>Acceleration</u> is the rate at which <u>velocity</u> changes.
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?