Explanation:
It is given that,
The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

Where,
x is in meters and t is in sec
We know that,
Velocity,

(a) i. t = 2 s

At t = 4 s

(b) Acceleration,

Pu t = 3 s in the above equation
So,

Hence, this is the required solution.
<span>1200 meters is less than 1 kilometer
</span>is false