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

Answer:

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

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 $s=ut+\frac{1}{2}at^2$

$7.79=11\times 2.55+\frac{1}{2}\times a\times 2.55^2$

$a=-6.2331cm/sec^2$