Question

a train starting from rest attains a velocity of 72 km per hour in 5 minute as human the acceleration is uniform find
1) the acceleration
2) the distance travelled by the train for attaining this velocity ​

Answer 1

Answer:

$\bigstar{\bold{Acceleration=0.067\:m/s^{2} }}$

$\bigstar{\bold{Distance\:travelled=2985\:m}}$

Explanation:

$\Large{\underline{\underline{\sf{Given:}}}}$

• Initial velocity (u) = 0 m/s
• Final velocity (v) = 72 km/hr = 20 m/s
• Time taken(t) = 5 min = 300 s

$\Large{\underline{\underline{\sf{To\:Find:}}}}$

• Acceleration
• Distance travelled

$\Large{\underline{\underline{\sf{Solution:}}}}$

Acceleration:

➜ First we have to find the acceleration of the train

➜ By the first equation of motion we know that,

    v = u + at

➜ Substitute the data,

    20 = 0 + a × 300

    a = 20/300

    a = 0.067 m/s²

➜ Hence acceleration of the train is 0.067 m/s²

    $\boxed{\bold{Acceleration=0.067\:m/s^{2} }}$

Distance travelled:

➜ Now we have to find the distance travelled by the train

➜ By the third equation of motion,

    v² - u² = 2as

➜ Substitute the data,

    20² - 0² = 2 × 0.067 × s

    400 = 0.134 × s

    s = 400/0.134

    s = 2985 m

➜ Hence distance travelled by the train is 2985 m

    $\boxed{\bold{Distance\:travelled=2985\:m}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

➜ The three equations of motion are :

• v = u + at
• s = ut + 1/2 × a × t²
• v² - u²= 2as