Question

A box slides down an incline with uniform acceleration. It starts from rest and attains a velocity of 2.7 m/s in 30s. Find the acceleration and the distance moved in the first 6.0s .

Answer 1

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,

`a=(v-u)/(t)`

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,

`\begin{gathered} a=(2.7-0)/(30) \\ a=0.09ms^(-2) \end{gathered}`

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,

`S=ut+(1)/(2)at^2`

Here t is 6 seconds,

Substituting the known values,

`\begin{gathered} S=0* t+(1)/(2)*0.09*6^2 \\ S=1.62\text{ m} \end{gathered}`

Thus, the distance traveled by the box in 6 seconds is 1.62 m.