An object of mass $$3\,kg$$  is at rest. If a force $$F = \left( {6{t^2}\hat i + 4t\hat j} \right)N$$    is applied on the object, then the velocity of the object at $$t = 3\,s$$  is

Solution :
According to Newton's 2^nd law, force applied on an object is equal to rate of change of momentum.
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,F = \frac{{dp}}{{dt}} \cr & {\text{or}}\,F = m\frac{{dv}}{{dt}}\,.......\left( {\text{i}} \right) \cr} $$
$${\text{Given,}}\,\,m = 3\;kg,t = 3\;s,F = \left( {6{t^2}\hat i + 4t\hat j} \right)N$$
Substituting these values in Eq. (i), we get
$$\eqalign{ & \left( {6{t^2}\hat i + 4t\hat j} \right) = 3\frac{{dv}}{{dt}} \cr & {\text{or}}\,\,dv = \frac{1}{3}\left( {6{t^2}\hat i + 4\hat j} \right)dt \cr} $$
Now, taking integration of both sides, we get
$$\eqalign{ & \int d v = \int_0^t {\frac{1}{3}} \left( {6{t^2}\hat i + 4t\hat j} \right)dt \cr & v = \frac{1}{3}\int_0^t {\left( {6{t^2}\hat i + 4t\hat j} \right)} dt \cr & {\text{but}}\,\,t = 3\;s\,\,\left( {{\text{given}}} \right) \cr & \therefore v = \frac{1}{3}\int_0^3 {\left( {6{t^2}\hat i + 4t\hat j} \right)} dt \cr & {\text{or}}\,\,v = \frac{1}{3}\left[ {\frac{{6{t^3}}}{3}\hat i + \frac{{4{t^2}}}{2}\hat j} \right]_0^3 \cr & {\text{or}}\,\,v = \frac{1}{3}\left[ {2{{\left( 3 \right)}^3}\hat i + 2{{\left( 3 \right)}^2}\hat j} \right] \cr & {\text{or}}\,\,v = \frac{1}{3}\left[ {54\hat i + 18\hat j} \right] \cr & {\text{or}}\,\,v = 18\hat i + 6\widehat j \cr} $$