Question

A body at rest starts sliding from top of a smooth inclined plane and requires $4\,\sec $ to reach bottom. How much time does it take, starting from rest at top, to cover one-fourth of a distance?
A. $1\sec $
B. $2\sec $
C. $3\sec $
D. $4\sec $

Answer 1

Hint: Here in the given question there is only constant acceleration on the body, not having any external force on the body. Therefore we can apply the equation of motion. We have to choose from the three equations of motion to get the required value.

Complete answer:
Let acceleration of the object be $a$ and length of the incline plane $s$ .
Here the acceleration is constant, therefore we can apply the equation of motion here.
Now, according to second equation of motion,
$${\text{s = ut + }}\dfrac{1}{2}{\text{a}}{{\text{t}}^2}\,\,\,\,\,\,.......(i)$$
Here, ${\text{l}}$ is displacement, ${\text{u}}$ is initial velocity, ${\text{t}}$ is the time and ${\text{a}}$ is acceleration.
According to the given question, initial velocity of the object is zero,
$ \Rightarrow {\text{u = 0}}$
Therefore, $(i)$ becomes
$${\text{s = }}\dfrac{1}{2}{\text{a}}{{\text{t}}^2}\,\,\,\,\,\,.......(ii)$$
Now, putting the value of ${\text{t = 4sec}}$ , we get
$${\text{s = 8a}}\,\,\,\,\,\,.......(iii)$$
To find the time taken for one-fourth of distance travelled equation $(i)$ becomes,
$$\dfrac{{\text{s}}}{4}{\text{ = }}\dfrac{1}{2}{\text{a}}{{\text{t}}_1}^2\,\,\,\,\,\,.......(iv)$$
Now from equation $(iii)$ and $(iv)$, we get,
${{\text{t}}_1} = 2\sec $
Therefore, the object will take $2\sec $ to reach one-fourth of distance and option B is correct.

Note: We can apply the equations of motion where we have problems with constant acceleration. These equations are only valid for constant acceleration. Constant acceleration means change in velocity is constant with time.