Question

An object of mass 3kg is at rest. Now a force F=6t^2î+2t j N is applied on the object. Find the velocity of the object at t=3sec

Answer 1

Answer:

Velcity of a object at t=3sec is 21 m/s.

Explanation:

F = m $\frac{dv}{dt}$ = $6t^{2}$ i  +2t j

→ $\frac{dv}{dt}$ = ( $6t^{2}$ i  +2t j )/3 = $2t^{2}$ i + (2/3)t j

→ $dv$ = [ $2t^{2}$ i + (2/3)t j ] $dt$

Taking integration both sides,

$\int\limits^v_0 \ dv = \int\limits^3_0 {2t^{2} } \, dx$

$\int\limits^v_0 \ dv = \int\limits^3_0 {[ 2t^{2} \uvec{i} + (2/3)t j] } \, dt$

→ v = [ $\frac{2t^{3} }{3}$ + $\frac{2}{3}$ * $\frac{t^{2} }{2}$] at t = 3 sec

→ v = [18 + 3]= 21 m/s

Answer 2

Answer:

Velocity of the object at 3s is $18 \hat{i} +3 \hat{j} m/s$

Explanation:

Given mass of an object, $m=3kg$

         force applied on the object, $F=6t^{2} \hat{i} +2t \hat{j} N$

         time, $t=3sec$

From the relation of force, $F=ma$

acceleration is given by $a=\frac{F}{m}$

                                          $=\frac{6t^{2} \hat{i} +2t \hat{j} }{3}=2t^{2} \hat{i} +\frac{2t \hat{j}}{3}$

Acceleration is defined as rate of change of velocity, i.e.,

                                $a=\frac{dv}{dt}$

                       $\implies \frac{dv}{dt}=2t^{2} \hat{i} +\frac{2t \hat{j}}{3}$

                       $\implies {dv}=2t^{2} \hat{i} +\frac{2t \hat{j}}{3}dt$

Integrating both sides, between the time 0 to 3s,

                    $\implies \int {dv}=\int\limits^3_0 (2t^{2} \hat{i} +\frac{2t \hat{j}}{3})dt$

                        $\implies v=\frac{2t^{3}}{3} \hat{i} \Big|\limits^3_0+\frac{2t^{2} \hat{j}}{6}) \Big|\limits^3_0$

                                   $=\frac{2\times 3^{3}}{3} \hat{i} +\frac{2\times 3^{2} }{6} \hat{j}$

                                   $=18 \hat{i} +3 \hat{j} m/s$

Therefore, the velocity of the object at 3s is $18 \hat{i} +3 \hat{j} m/s$