Given data:
* The initial velocity of the box is 0 m/s.
* The final velocity of the box is 2.7 m/s.
* The time taken by the box to reach the final velocity is 30 s.
Solution:
(a). By the kinematics equation, acceleration of the box is,
[tex]a=\frac{v-u}{t}[/tex]where v is the final velocity, u is the initial velocity, t is the time taken, and a is the accelration,
Substituting the known values,
[tex]\begin{gathered} a=\frac{2.7-0}{30} \\ a=0.09ms^{-2} \end{gathered}[/tex]Thus, the acceleration of the box is 0.09 meter per second squared.
(b). As the acceleration remain uniform,
By the kinematics equation, the distance moved by the box in 6 seconds is,
[tex]S=ut+\frac{1}{2}at^2[/tex]Here t is 6 seconds,
Substituting the known values,
[tex]\begin{gathered} S=0\times t+\frac{1}{2}\times0.09\times6^2 \\ S=1.62\text{ m} \end{gathered}[/tex]Thus, the distance traveled by the box in 6 seconds is 1.62 m.
