A train starting from rest attins a velocity of 72km/h in 5minute. Assuming that the acceleration in uniform find a. The acceleration and b.the distance travelled by the train while it attined this velocity
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72 kmph = 72 × (5/18) m/s = 20 m/s
a = (v - u) / t
= [(20 - 0) m/s] / (5 × 60 s)
= 0.066 m/s²
Acceleration of train is 0.066 m/s²
S = ut + 0.5at²
= 0 + [0.5 × (20/300 m/s²) × (5 × 60 s)²]
= 3000 m
= 3 km
Distance travelled by the train during this time is 3 km
a = (v - u) / t
= [(20 - 0) m/s] / (5 × 60 s)
= 0.066 m/s²
Acceleration of train is 0.066 m/s²
S = ut + 0.5at²
= 0 + [0.5 × (20/300 m/s²) × (5 × 60 s)²]
= 3000 m
= 3 km
Distance travelled by the train during this time is 3 km
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