A train starting from rest attains a velocity of 36 km/h with 5minutes

1.	A train starting from rest attains a velocity of 36 km/h with 5minutes. Caluculate the distance to attain the velocity
Answer» Given:- Initial Velocity (u) = 0m/s Final Velocity (v) = 36km/h = 10m/s Time Taken = 5min = 300s. To FIND:- The Distance train travelled to attain the final Velocity. Formulae used:- v = u + at v² - u² = 2as. Where, v = Final Velocity u = Initial Velocity a = Acceleration s = Distance t = Time. Now, Using first equation of motion to find accceleration. v = u + at or, a = v - u/t a = 10 - 0 / 300 a = 10/300 a = 0.03m/s² Hence, The Acceleration is 0.03m/s². Therefore, using Third equation of motion to find Distance travelled. v² - u² = 2as (10)² - (0)² = 2 × 0.03 × s 100 = 0.06 × s S = 100/0.06 S = 1.666km. Hence, The distance travelled by train is 1.6km to attain the Velocity.

A train starting from rest attains a velocity of 36 km/h with 5minutes. Caluculate the distance to attain the velocity

Answer»

Given:-

Initial Velocity (u) = 0m/s

Final Velocity (v) = 36km/h = 10m/s

Time Taken = 5min = 300s.

To FIND:-

The Distance train travelled to attain the final Velocity.

Formulae used:-

v = u + at

v² - u² = 2as.

Where,

v = Final Velocity

u = Initial Velocity

a = Acceleration

s = Distance

t = Time.

Now, Using first equation of motion to find accceleration.

v = u + at

or,

a = v - u/t

a = 10 - 0 / 300

a = 10/300

a = 0.03m/s²

Hence, The Acceleration is 0.03m/s².

Therefore, using Third equation of motion to find Distance travelled.

v² - u² = 2as

(10)² - (0)² = 2 × 0.03 × s

100 = 0.06 × s

S = 100/0.06

S = 1.666km.

Hence, The distance travelled by train is 1.6km to attain the Velocity.

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