Answer
(a) $76.5\space N.s$
(b) $546.4 N$
(c) $31.5m/s$
Work Step by Step
(a) We can find the required impulse as follows:
$I=\frac{1}{2}F_{avg}\Delta t$
As the area of trapezium is given as
$\Delta t=A=\frac{1}{2}(a+b)h$
Now $I=\frac{1}{2}(0.03+0.14)(900)$
$\implies I=76.5\space N.s$
(b) The average force can be determined as
$F_{avg}=\frac{I}{\Delta t}$
We plug in the known values to obtain:
$F_{avg}=\frac{76.5}{0.14}$
$\implies F_{avg}=546.4N$
(c) We can find the required speed as follows:
$v_f=\frac{I}{m}+v_i$
We plug in the known values to obtain:
$v_f=\frac{76.5}{3}+6$
$\implies v_f=31.5m/s$
