an object moving with uniform acceleration has a velocity of 11.0 cm/s in the positive x-direction when its x-coordinate is 2.91 cm. if its x-coordinate 3.25 s later is −5.00 cm, what is its acceleration?

Respuesta :

The acceleration of the object is 5.3 cm/[tex]s^{2}[/tex] and is directed towards the negative x - axis.

We have an object moving with uniform acceleration along the + x axis.

We have to determine of the acceleration of object after 3.25 seconds.

According to the question -

initial velocity = u = 11 cm/s along +x axis.

Now, using the second equation of motion -

[tex]$S=ut +\frac{1}{2}at^{2}[/tex]

S = x(2) - x(1) = - 5 - 2.91 = - 7.91 = 7.91 cm along -x axis.

u = 11 cm/s

Substituting the values -

7.91 = 11 x 3.25 + 0.5 x 3.25 x 3.25  x a

7.91 = 35.75 + 5.3a

a = - 5.3 cm/[tex]s^{2}[/tex]

Hence, the acceleration of the object is 5.3 cm/[tex]s^{2}[/tex] and is directed towards the negative x - axis.

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