Question

Over a time interval of 1.71 years, the velocity of a planet orbiting a distant star reverses direction, changing from +17.3 km/s to -22.8 km/s. Find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s2) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.

Answer 1

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