Question

A body of mass 0.40 kg moving initially with a constant speed of 10 m 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

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 =$\frac{F}{m}$ = $\frac{-8.0}{0.40}$ = -20 $m/s^2$
At t = –5 s
Acceleration, $a$ = 0 and $u$ = 10 $m/s$

s = ut + a't2

s = 10 × (–5) = –50 m
At t = 25 s
Acceleration, $a''$= – 20 $m/s^2$ and $u$ = 10 $m/s$

$s$1 = $ut$ + $\frac{1}{2}$a''t2

= 10x25 + $\frac{1}{2}$ x (-20) x (25)2
= 250 +6250 = -6000 m
At t = 100 s
For 0 $\leq$ t $\leq$ 30 s
a = –20 $m/s^2$
u = 10 $m/s$

$s_1$ = $ut+\frac{1}{2}at^2$
= 10 x 30 + $\frac{1}{2}$ x (-20) x (30)2
= 300 - 9000
= - 8700 m
For 30'< t $\leq$ 100 s
As per the first equation of motion, for t = 30 s, final velocity is given as:
v = u + at
= 10 + (–20) × 30 = –590 $m/s$
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''t2

= –590 × 70 = – 41300 $m$
∴ Total distance, $s$‘’ = $s_1+s_2$ = - 8700 - 41300 = - 50000 $m$