Given,
Body mass is 0.40 kg.
u = 10 m/s initial velocity
f = -8 N force (retarding force)
Using the formula S = ut + (12) at2,
(a) At time t = – 5 s, position
From t = 0 s, the force acts on the body.
As a result, when the time is – 5 s, the body’s acceleration is 0.
S1= (10)(-5) + (½) (0) (-5)2 = – 50 m
(b) At time t = 25 s, position
The force acting in the opposite direction causes the body to accelerate.
a = F/a = -8 /0.4 = -20 ms-2
S2= (10)(25) + (½) (-20) (25)2 = – 6000 m
2 = − 6000 m
(c) At time t = 100 s, position
The body will move under the force’s retardation for the first 30 seconds, then the speed will remain constant.
As a result, distance covered in 30 sec
S3= (10)(30) + (½)(-20)(302)
= 300 – 9000 = – 8700m
The speed after 30 sec is
v = u + at
v = 10 – (20 x 30) = 590 m/s
The distance covered in the next 70 sec is
S4 = – 590 x 70 + (½) (0) (70)2 = – 41300 m
Therefore the position after 100 sec = S3 + S4 = – 8700 – 41300 = – 50000m