Question

A body of mass 0.40 kg moving initially with a constant speed of $10m{s}^{-1}$ to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = -5 s, 25 s, 100 s.

Answer 1

Given:
Mass of the body, m = 0.40 kg
Initial speed of the body, u = 10 m/s due north
Force acting on the body, F = -8.0 N
Acceleration produced in the body, $a=\frac{F}{m}=\frac{-8}{0.40}=-20m/s^2$
At t= -5s:
Acceleration, a' = 0 and u = 10 m/s
$s=ut+\frac{1}{2}{a}^{\prime }{t}^2=10\times (-5)=-50m$

At t = 25 s
Acceleration, $a"=-20m/s^{2}andu=10m/s{s}^{\prime }=u{t}^{\prime }+\frac{1}{2}a"t^{\prime 2}=10\times 25+\frac{1}{2}\times (-20)\times (25{)}^2=250-6250=-6000m$

At t = 100 s
For $0\le t\le 30s{a}^{\prime \prime }=-20m/s^{2}u=10m/s{s}_1=ut+{\frac{1}{2}}^{\prime \prime }{t}^2=10\times 30+\frac{1}{2}\times (-20)\times (30{)}^2=300-9000=-8700m$
For $30<t\le 100s$
As per the first equation of motion, for t = 30 s, final velocity is given as:
$v=u+at=10+(-20)\times 30=-590m/s$
Velocity of the body after 30 s = -590 m/s
For motion between 30 s to 100 s, i.e., in 70 s:
$s_2=vt+\frac{1}{2}a"t^2=-590\times 70=-41300m\therefore Totaldistance,s"=s_1+s_2=-8700-41300=-50000m=-50km$