Question

The velocity of a particle is $v=v_o+gt+F{t}^2$. Its position is $x=0$at $t=0$; then its displacement after time $\left(t=1\right)$ is

• A. $v_o+\frac{g}{2}+F$
• B. $v_o+2g+3F$
• C. $v_o+g+F$
• D. $v_o+\frac{g}{2}+\frac{F}{3}$

Answer 1

The correct option is D

$v_o+\frac{g}{2}+\frac{F}{3}$

Step 1: Given data:

The velocity of a particle is $v=v_o+gt+F{t}^2\dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

Condition; $x=0$at $t=0$

Step 2: Formula used:

Velocity is defined as the rate of change of displacement that is,

$v=\frac{dx}{dt}\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(2\right)$

Where, $v,x,t$ are velocity, displacement, and time respectively.

Step 3: Calculation of displacement equation:

We know that, from equation (2),

$v=\frac{dx}{dt}$

It can be written as-

$dx=vdt$

substitute the value of velocity from equation (1) and then integrating both the sides

$\begin{array}{rcl}dx& =& \left(v_o+gt+F{t}^2\right)dt\\ \int dx& =& \int \left(v_o+gt+F{t}^2\right)dt\\ x& =& v_{o}t+g\frac{t^2}{2}+F\frac{t^3}{3}+c\dots \dots \dots \dots \dots \dots \dots \dots \left(3\right)\end{array}$

$c=cons\tan t$

Now from the given conditions,

$x=0$at $t=0$

Putting this condition in equation (3),

$$\begin{gathered} 0=0+0+0+c \\ c=0 \end{gathered}$$

Put the value of constant in equation (3) we get,

$x=v_{o}t+\frac{g{t}^2}{2}+\frac{F{t}^3}{3}\dots \dots \dots \dots \dots \dots \dots \left(4\right)$

The displacement equation is $x=v_{o}t+\frac{g{t}^2}{2}+\frac{F{t}^3}{3}$.

Step 4: Calculation of displacement after time $\left(t=1\right)$:

For obtaining the displacement after time $\left(t=1\right)$ put $\left(t=1\right)$ in equation (4) we get-

$$\begin{gathered} x=v_{o}t+\frac{g{t}^2}{2}+\frac{F{t}^3}{3} \\ x=v_o\left(1\right)+\frac{g\left(1^2\right)}{2}+\frac{F\left(1^3\right)}{3} \\ x=v_o+\frac{g}{2}+\frac{F}{3} \end{gathered}$$

Thus, its displacement after time $\left(t=1\right)$ is $v_o+\frac{g}{2}+\frac{F}{3}$.

Hence, option (D) is the correct answer.