Given,
u = 0
t = 10 s
a = 2 m s-2
v = u + at
Substituting the values in the formula above we get,
v = u + at = 0 + 2 × 10 = 20 m s-1
Hence, final velocity = 20 m s-1
(ii) Retardation in last 50 s,
v = u + at
Given,
u = 20 m s-1
v = 0 m s-1
t = 50 s
Substituting the values in the formula above we get,
0 = 20 + (-a) × 50
or, 50 a = 20
or, a = 20/50 = 0.4 m s-2
Hence, retardation in the last 50 s = 0.4 m s-2
(iii) As we know,
S = ut + 1/2 at2
For the first 10 s,
u = 0
a = 2 ms-2
S1 = (0 × 10) + 1/2 × 2 ×102
= 0 + 102 = 100 m = 100 m
For the next 200 s,
S2 = speed x time
where speed = 20
time = 200 s
Substituting the values in the formula above we get,
S2 = 20 × 200 = 4000 m
For the next 50 s
S3 = (20 × 50) + 1/2 × (-0.4) × 50²
= 1000 – 0.2 × 2500
= 1000 – 500 = 500 m
So, Total distance = S1 + S2 + S3
= 100 m + 4000 m + 500 m
= 4600 m
Hence, total distance covered = 4600 m
(iv) Average velocity = 4600/260
= 17.69 m s-1
Hence, average velocity = 17.69 m s-1