Answer to Question #126905 in Physics for Asif Sultan

Question

http://prntscr.com/tkmq8j
Explain the solution?
Thanks

Answer 1

(1) According to the condition of the problem

ans. (see fig.1) $a=(\overrightarrow{F}_2+\overrightarrow{F}_3)/m$

(2) According to the schedule the force is changed

$a=dv/dt$ and $dv/dt=F/m\to dv=(F/m)dt\to v=\int{(F/m)dt}$

1) $F=2t$ $\to v_1=7m/s$

2) $F=4$ $\to v_2=15m/s$

3) $F=-3t+16$ $\to v_3=22.625m/s$

4) $F=2.5$ $\to v_4=27.625m/s$ Answer.

Fig.1