Abody moving with uniform acceleration has a velocity of 15.3 cm/s when its x coordinate is 4.3 cm. if its x coordinate 1.72 s later is −2.32 cm, what is the x-component of its acceleration? answer in units of cm/s 2 .

a=-22.266cm/s^2

Explanation:

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} =ΔX(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

X=-2.32cm

Xo=4.3cm

Vo=15.3cm/s

t=1.72s

WE USE THE ECUATION 3

a=(X-Xo-Vot)2/t^2

a=(-2.32-4.3-15.3x1.72)2/1.72^2

a=-22.266cm/s^2

it is 33

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