Question

A body starts from rest with uniform acceleration. The velocity of the body after t seconds is v. The displacement of the body in last three seconds is

• A. $\frac{3v}{2\left(t-3\right)}$
• B. $\frac{3v}{2\left(t+3\right)}$
• C. $3v\left(1-\frac{3}{2t}\right)$
• D. $3v\left(1+\frac{3}{2t}\right)$

Answer 1

The correct option is C

$3v\left(1-\frac{3}{2t}\right)$

Step 1: Given data

1. The velocity of the body at the time of t seconds is v.
2. The initial velocity of the body is zero.
3. acceleration of the body is constant.

Step 2: Displacement

1. Displacement is a vector quantity. The distance travelled by a body without changing its direction is called displacement.
2. The displacement formula of kinematics is, $S=ut+\frac{1}{2}a{t}^2$ , where s and u are the distance travelled by the body and initial velocity respectively, a is the acceleration and t is the time.
3. Again $s=vt$, v is the velocity of a body.

Step 3: Finding the displacement

consider the displacement of the body is s at time t.

Displacement of the body at time t is

$s_1=\left(0\times t\right)+\frac{1}{2}a{t}^2$

or $s_1=\frac{1}{2}a{t}^2$ ……………..(1).

Displacement of the body at time (t-3) is,

$s_2=\frac{1}{2}a\times {\left(t-3\right)}^2$ …………….(2)

Now, displacement of the body in the last three seconds is from time (t-3) to t seconds,

$s=s_1-s_2=s=\frac{1}{2}a{t}^2-\frac{1}{2}a{\left(t-3\right)}^2$

$$\begin{gathered} s=\frac{1}{2}a\left[t^2-{\left(t-3\right)}^2\right] \\ ors=\frac{1}{2}a\left[t^2-\left(t^2-6t+9\right)\right] \\ ors=\frac{1}{2}a\left(t^2-t^2+6t-9\right) \\ ors=\frac{1}{2}a\left(6t-9\right)............\left(3\right) \end{gathered}$$

Again, acceleration, $a=\frac{v}{t}$,

So, from the equation (3),

$$\begin{gathered} s=\frac{1}{2}\times \frac{v}{t}\times \left(6t-9\right) \\ ors=\frac{v}{t}\times \left(3t-\frac{9}{2}\right) \\ ors=3v-\frac{9v}{2t} \\ ors=3v\left(1-\frac{3}{2t}\right) \end{gathered}$$

So, the displacement of the body in the last three seconds is $3v\left(1-\frac{3}{2t}\right)$. Therefore option (c) is correct.