Answer:
c = 4,444.44
Explanation:
You have the following expression for the acceleration of the projectile:
(1)
s: distance to the ground of the projectile
To find the value of the constant c you use the following formula:
(2)
vo: initial velocity = 0 m/s
v: final speed = 200 m/s
Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m
You replace the expression (1) into the expression (2):
You do the constant c in the last equation, then you replace the values of v, s and Δs:
Answer:
The car that is accelerating is B a car that rounds a curve at a constant speed
Explanation:
Although all of the cars are at a constant speed or not moving acceleration is the change in speed or the change of directions therefore making the only car changing directions your answer.
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:
wA - T =mA(a)
second option
Explanation:
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