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?
Answer: The amount of energy consisted in the molecules determines the state of matter.
Explanation:
Answer:
Acceleration=24.9ft^2/s^2
Angular acceleration=1.47rads/s
Explanation:
Note before the ladder is inclined at 30° to the horizontal with a length of 16ft
Hence angular velocity = 6/8=0.75rad/s
acceleration Ab=Aa +(Ab/a)+(Ab/a)t
4+0.75^2*16+a*16
0=0.75^2*16cos30°-a*16sin30°---1
Ab=0+0.75^2sin30°+a*16cos30°----2
Solving equation 1
(0.75^2*16cos30/16sin30)=angular acceleration=a=1.47rad/s
Also from equation 2
Ab=0.75^2*16sin30+1.47*16cos30=24.9ft^2/s^2
first is gamma Ray's, last is d
Answer:
d= 64.7 km

displacement vector
Explanation:
total distance = 40 + 30 = 70 km
during 1st flight


during 2nd flight



the two component of r are:


Geographical Direction ![\theta = tan^{-1}\frac{r_y}{r_x} [tex]\theta = 40.9^{o}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%5Cfrac%7Br_y%7D%7Br_x%7D%20%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Ctheta%20%3D%2040.9%5E%7Bo%7D)
Displacement d

d= 64.7 km
displacement vector