A train starts from rest and accelerates uniformly at a rate of 2 m s-2 for 10 s. It then maintains a constant speed for 200 s. The brakes are then applied and the train is uniformly retarded and comes to rest in 50 s. Find — (i) the maximum velocity reached, (ii) the retardation in the last 50 s, (iii) the total distance travelled, and (iv)the average velocity of the train.

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