A train starting from rest attains a maximum speed of 54km/h in 30 s with uniform acceleration, it then travel with this speed for 40 s and is then brought to rest in further 30 s with uniform retardation, calculate the total distance travelled.
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A train starting from rest attains a maximum speed of 54km/h = 54 × 5/18 = 15 m/s with uniform acceleration.
use formula, v = u + at
here, u = 0, v = 15m/s , t = 30s
so, 15 = 0 + 30a => a = 0.5 m/s²
so, distance covered during motion , S = ut + 1/2at²
= 0 + 1/2 × 0.5 × (30)² = 225 m
then, train with maximum speed 54km/h or 15m/s for 40s.
so, distance covered during this motion , S' = vt
= 15 × 40 = 600m
now, train is brought to rest further 30sec with uniform retardation . so, retardation = -0.5 m/s² [ magnitude of retardation is same acceleration this is due to time taken during motion is same.]
so, distance covered , S" = ut + 1/2at²
S" = 15 × 30 - 1/2 × 0.5 × (30)²
= 450 - 225 = 225m
hence, total distance covered by train = 225 + 600 + 225 = 1050m
use formula, v = u + at
here, u = 0, v = 15m/s , t = 30s
so, 15 = 0 + 30a => a = 0.5 m/s²
so, distance covered during motion , S = ut + 1/2at²
= 0 + 1/2 × 0.5 × (30)² = 225 m
then, train with maximum speed 54km/h or 15m/s for 40s.
so, distance covered during this motion , S' = vt
= 15 × 40 = 600m
now, train is brought to rest further 30sec with uniform retardation . so, retardation = -0.5 m/s² [ magnitude of retardation is same acceleration this is due to time taken during motion is same.]
so, distance covered , S" = ut + 1/2at²
S" = 15 × 30 - 1/2 × 0.5 × (30)²
= 450 - 225 = 225m
hence, total distance covered by train = 225 + 600 + 225 = 1050m
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Answer:
1050 m
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