A constant resistive force starts acting on body of mass 10 kg moving with speed of 10m/s.If 5 sec after the f

Question

A constant resistive force starts acting on body of mass 10 kg moving with speed of 10m/s.If 5 sec after the force starts acting on the body its speed is 8m/s.After how much time,fromthis instant,will body comes to rest

"give calculation process"

Answer 1

Using v = u + at since the force is constant

First 5 seconds

8 = 10 + 5a

a = -2/5

Rest of motion, assuming same force

0 = 8 + (-2/5)t

t = 20 seconds

Answer 2

FOR THE EVENT LASTING FOR THE FIRST 5 SECONDS....
....v=FINAL VELOCITY=8m/s
....u=initial velocity=10m/s..
...time taken=t=5sec...
..so using v=u+at.....where a=acceleration(here negative as the force is resistive...so either consider a with sign or chose it without sign and get a negative answer...and take the negative sign to mean that aresistive force is acting,which infact is)..
....=>8=10+5a..
...therefore...a=-0.4m/s^2..
...now consider an event beginning when speed becomes 8m/s and ending when speed becomes zero..let it last for T seconds.......
...new final velocity=v1=0..
...new initial velocity=u1=v=8m/s...
....acceleration=-0.4m/s^2..
...again using 1st equation of motion for this event..
...final velocity=initial velocity + (acceleration*time)...
....v1=u1+aT...
....solving.....T=20 seconds...
...

Answer 3

Relative velocity is 8 - 10 = -2 m/s

a = v/t = -2/5 = -0.4 m/s^2

t = a/v = 8/0.4 = 20 sec