Question

Initial speed of the train, u = 90 km/h = 90 x 1000 = 25 ms-1

(60 x 60)

Final speed of the train, v = 0 (finally the train comes to rest) Acceleration = - 0.5 m s-2

According to third equation of motion:

2as = v2

- u2

s = (v2

- u2

) / 2a

s = (0 - 252

) / 2 x (-0.5) = 625 m

The train will cover a distance of 625 m before it comes to rest.

Ans: 3) Given

Initial velocity of the trolley, u = 0

Acceleration, a = 2 cm s-2 = 2 /100 = 0.02 m/s2 Time, t = 3s

According to the first equation of motion:

v = u + at

v = 0 + 0.02 × 3 = 0.06 m/s

Hence, the velocity of the trolley after 3s from start is 0.06 m/s.​

Answer 1

Answer:

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