(a) From O to A the body is said to have uniform acceleration and from A to B the body is said to have zero acceleration.
(b) Along OA the speed is increasing uniformly with respect to time and along AB the speed remains constant.

Speed of the train = 90 kmph
Speed of the train = \(\frac { 90000 }{ 3600 } \)
Speed of the train =\(\frac { 900 }{ 36 } \)=25
Speed of the train = 25 m/s.
V = 25 m/s
a = - 0.5 ms^-2
Displacement = \(\frac { V }{ a } =\frac { 25 }{ 0.5 } =\frac { 25 }{ -0.5 } \) = 50 seconds
S = ut + \(\frac { 1 }{ 2 } \)at²
= 25 x 50 - \(\frac { 1 }{ 2 } \) x 0.5 x 50² = 1250 - 625
Distance covered by the train before it is brought to rest = 625 m/s.

Speed of the signal = Speed of the light = 3 x 10⁸ ms^-1
Time taken by the signal to reach the ground = 5 seconds
Speed = \(\frac { Distance }{ time\ taken } \)
3 x 10⁸ ms^-1 =\(\frac { Distance }{ 5s } \)
Distance = 3 x 10⁸ x 5
= 15 x 10⁸ m
Distance of the spaceship from the ground station = 15 x 10⁸ m.

Track diameter \((\mathrm{d})=200 \mathrm{~m} ; \quad \mathrm{r}=\frac{200}{2}=100 \mathrm{~m}\)
Time taken for one rotation = 40 s
Total distance of the track \(=2 \pi r\)
\(=2 \times 3.14 \times 100 \)
\(=2 \times 314 \)
\(=628 \mathrm{~m}.\)
Displacement at the end of 2 m and 20 s =(2 x 60) + 20
=140 second
40 second =1 rotation
140 second \(=\frac{140}{40}=3.5\) rotation.
distance covered = 628 x 3.5
= 2198 m
displacement = 200 m