Question

An object moving with uniform acceleration has a velocity of 11.0 cm/s in the positive x-direction when its x-coordinate is 2.79 cm. If its x-coordinate 2.55s later is -5.00cm, what is its acceleration?

Answer 1

Acceleration will be
`a=-6.2331cm/sec^2`

Step-by-step explanation:

We have given speed of the moving object u = 11 cm /sec

As the object move from +2.79 x -coordinate to -5 x- coordinate

So total distance = 5+2.79 = 7.79 cm

Time is given as t = 2.55 sec

We have to find the acceleration

From second equation of motion we know that
`s=ut+(1)/(2)at^2`

`7.79=11* 2.55+(1)/(2)* a* 2.55^2`

`a=-6.2331cm/sec^2`