An object is moving with an initial velocity of 23 m/s. it is then subject to a constant acceleration of 3.5 m/s2 for 12 s. how far will it have traveled during the time of its acceleration?

528 m

Explanation:

We can solve the problem by using the following suvat equation:

where

s is the displacement

u is the initial velocity

t is the time

a is the acceleration

In this problem,

u = 23 m/s

And substituting t = 12 s, we find the distance travelled:

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.