Question

An object of mass three kilogram is at rest. Now a force $\overrightarrow {\text{F}} = 6{t^2}\mathop i\limits^ \wedge + 4t\mathop j\limits^ \wedge $ is applied on the object then velocity of the object at ${\text{t = 3}}$ second is.
A. ${\text{18}}\mathop i\limits^ \wedge + 3\mathop j\limits^ \wedge $
B. ${\text{18}}\mathop i\limits^ \wedge + 6\mathop j\limits^ \wedge $
C. $3\mathop i\limits^ \wedge + 18\mathop j\limits^ \wedge $
D. ${\text{18}}\mathop i\limits^ \wedge + 4\mathop j\limits^ \wedge $

Answer 1

Hint: In this question a force of $6{t^2}\mathop i\limits^ \wedge + 4t\mathop j\limits^ \wedge$ is applied on the object therefore consider ${\overrightarrow {\text{F}} _x} = 6{{\text{t}}^2}\mathop {\text{i}}\limits^ \wedge $ and ${{\text{F}}_y} = 4t\mathop j\limits^ \wedge $.

Complete step-by-step answer:
Formula used: ${\text{Acceleration = }}\dfrac{{{\text{Force}}}}{{{\text{Mass}}}}$ i.e. $\overrightarrow {\text{a}} = \dfrac{{\overrightarrow {\text{F}} }}{{\text{m}}}$

Given that,

$\overrightarrow {\text{F}} = 6{t^2}\mathop i\limits^ \wedge + 4t\mathop j\limits^ \wedge $

Mass $ = 3{\text{kg}}$
Time $ = 3$ seconds
Therefore ${\overrightarrow {\text{F}} _x} = 6{t^2}\mathop i\limits^ \wedge $

As we know that $\overrightarrow {\text{a}} = \dfrac{{\overrightarrow {\text{F}} }}{{\text{m}}}$
so, ${{\text{a}}_x} = \dfrac{6}{3}{t^2}\mathop i\limits^ \wedge $
${{\text{a}}_x} = {\text{2}}{{\text{t}}^2}\mathop i\limits^ \wedge $
$\dfrac{{d{v_x}}}{{dt}}{\text{ = 2}}{{\text{t}}^2}\mathop i\limits^ \wedge $

integrating both sides we get:

$
  \int\limits_0^{{v_2}} {d{v_x} = \int\limits_0^3 {2{t^2}} dt} \\
  {V_x} = \dfrac{{ - 2}}{3}{t^3}\int\limits_0^3 { = 18{\text{m/s}}} \\
$

Now ${{\text{F}}_y} = 4t\mathop j\limits^ \wedge $
therefore ${{\text{a}}_y} = \dfrac{4}{3}t\mathop j\limits^ \wedge $

Integrating both sides we get:
$
  \int\limits_0^{{v_y}} {d{v_y} = \int\limits_0^3 {\dfrac{4}{3}t} dt} \\
  {V_y} = \dfrac{{{t^2}}}{3}\int\limits_0^3 { = 6{\text{m/s}}} \\
$

Therefore the velocity of object i.e. ${\text{V = 18}}\mathop i\limits^ \wedge + 6\mathop j\limits^ \wedge $

Hence the correct option is B.

Note: In this question the force towards both the axis is given along with mass which is three kilogram hence we calculated the acceleration using the formula after that integrating both the resulted equation from limit zero to three we calculated the velocity of object as ${\text{18}}\mathop i\limits^ \wedge + 6\mathop j\limits^ \wedge $.