Question

A body moving with uniform retardation covers 3 km before its speed is reduced to half of its initial value. It comes to rest in another distance of
A.) 1 km
B.) 2 km
C.) 3 km
D.) 1/ 2 km

Answer 1

Hint: Assume the initial velocity to be v m/s, so the final velocity will be v/ 2 and then use the equation of motion, ${v^2} - {u^2} = 2as$ , and then solve the question.

Step By Step Answer:

Formula used : ${V^2} - {u^2} = 2as$

We have been given in the question that-

A body moving with uniform retardation covers a distance of 3 km.
So, let us assume that the initial velocity is v.
As, given in the question that the speed reduces to half.
So, the final velocity will be v/ 2.

Now, using the equation of motion-

${V^2} - {u^2} = 2as$
Here, V is the final velocity, u is the initial velocity, a is the acceleration and S is the displacement.

Now putting, S = 3 km, u = v m/s, V = v/2 m/s, we get-
${\dfrac{v}{4}^2} - {v^2} = 2a \times 3$
Solving further, we get-
${\dfrac{{ - 3v}}{4}^2} = 6a \\\ \Rightarrow a = {\dfrac{{ - 3v}}{{24}}^2} = {\dfrac{{ - v}}{8}^2} - (1) \\\$
Now, also given in the question that the body comes to rest finally after travelling a certain distance.

So, in this case initial speed will be u = v/2 and the final speed will become V = 0, as the body comes to rest.
Let the distance covered be x cm, so we can write by again using the equation ${V^2} - {u^2} = 2as$
$0 - {\dfrac{v}{4}^2} = 2aX \\\ \Rightarrow {\dfrac{{ - v}}{4}^2} = 2aX \\\$
Putting the value of a from equation (1), we get-
${\dfrac{{ - v}}{4}^2} = 2 \times \dfrac{{ - {v^2}}}{8} \times X \\\ \Rightarrow {\dfrac{{ - v}}{4}^2} = {\dfrac{{ - v}}{4}^2} \times X \\\ \Rightarrow X = 1 \\\$

So, the body comes to rest at another distance of 1 km.

Hence, the correct option is A.

Note: Whenever solving such types of questions, always write down the things given in the question. And then as mentioned in the solution, using the equation of motion, and putting the value of final and initial velocity and distance as 3 km, we found the value of a (acceleration) in terms of v and then again by putting the initial velocity as v/2 and final as 0, we found out the distance travelled.