Answer:
- R = ( 4.831 m , 1.469 m )
- Direction of R relative to the x axis= 16°54'33'
Explanation:
Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula

where
its the magnitude and θ.
So, for our vectors, we will have:


and


Now, we can take the sum of the vectors




This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem





To find the direction, we can use




As we are in the first quadrant, this is relative to the x axis.
Answer:
The si unit to measure force are Pascels or Pa
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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?