Answer:
a)40100m/s
b)-4.348x10^- m/s^2
Explanation:
to calculate the change in the planet's velocity we have to rest the speeds
ΔV=-22.8-17.3=-40.1km/s=40100m/s
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem we have to convert the time interval ins seconds, we know that a year has 53926560s
t=1.71years=53926560*1.71=92214417.6
then we can use the ecuation number 1 to calculate the aceleration
Vf=-22.8km/s
Vo=17.3km/s
Vf=Vo+at
a=(vf-vo)/t
a=(-22.8-17.3)/92214417.6
a=-4.348x10^-7 km/s^2=-4.348x10^- m/s^2
Answer:
they will move away from each other
Answer: the maximum heigth of the stadium at ist back wall is 151.32 ft
Explanation:
1. use the position (x) equation in parobolic movement to find the time (t)
565 ft = [frac{176 ft}{1 s\\}[/tex] * cos (35°) * t
t= 3.92 s
2. use the position (y) equation in parabolic movement to find de maximun heigth the ball reaches at 565 ft from the home plate.
y= [[frac{176 ft}{1 s\\}[/tex] * sen (35°) * 3.92 s] - 
y= 148.32 ft
3. finally add the 3 ft that exist between the home plate and the ball
148.32 ft + 3 ft = 151.32
