Question

a train starting from rest attains a velocity of 72 km/hr in 5 min . assuming that the acceleration is uniform find the
1) acceleration
2) distance traveled by the train for attaining this velocity.




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Answer 1

$\huge\sf\pink{Answer}$

☞ Acceleration of the body is 0.06 m/s & Distance is 3333.34 m

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$\huge\sf\blue{Given}$

✭ Initial Velocity (u) = 0/s

✭ Final Velocity (v) = 72 km/h

✭ Time (t) = 5 min

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$\huge\sf\gray{To \:Find}$

◈ Acceleration & Distance travelled?

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$\huge\sf\purple{Steps}$

So before solving we shall first convert km/h to m/s and minutes to seconds

✏ Final Velocity = 72 km/h

➢ $\sf 72\times \dfrac{5}{18}$

➢ $\sf \green{v = 20 \ m/s}$

✏ Time = 5 min

➢ $\sf 5 \times 60$

➢ $\sf \green{t = 300 \sec}$

Now the acceleration of the body can be found with the help of the first equation of the motion,i.e

$\sf \underline{\boxed{\sf v = u+at }}$

Substituting the values,

➝ $\sf 20 = 0+a\times 300$

➝ $\sf 20=300a$

➝ $\sf \dfrac{20}{300} = a$

➝ $\sf \red{Acceleration = 0.06 \ m/s}$

Now the distance travelled with the help of the third equation of motion,i.e

$\sf \underline{\boxed{\sf v^2-u^2 = 2as}}$

➳ $\sf 20^2-0^2 = 2\times 0.06\times s$

➳ $\sf 400 = 0.12s$

➳ $\sf \dfrac{400}{0.12} = s$

➳ $\sf\orange{Distance = 3333.34m \ or \ 3.33334 \ km}$

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Answer 2

It is given that the initial position of the train was stationary. Then, it attains a velocity of 72 km/hr. (20 m/s) (v) in just 5 minutes (t).

As per the question, we have to find the acceleration (a) of the particle (train) assuming it's acceleration constant and distance (s) travelled in that certain period of time.

We know,

Acceleration = Rate of change of velocity

Or, (mathematically)

a = (v-u)/t

where, a is the acceleration, v and u are the velocities (final - initial) and t is the time taken.

Thus,

a = (20 m/s - 0)/(5 × 60 s)

=> a = (20 m/s)/(300 s)

=> a = (2 m/s)/(30 s)

=> a = (1 m/s)/(15 s)

=> a = 0.0666666666... m/s²

=> a = 0.06 m/s² (Approximately)

Now, fetching the value of (s):-

We know,

s = ut + 1/2at^2

where, s is the distance travelled, t is time take, a is acceleration, u is initial velocity.

Hence,

s = 1/2at^2

=> s = (1/2)(0.06 m/s²) × (300 s)^2

=> s = (0.03 m/s²)(90000 s²)

=> s = (3 m/s²)(900 s²)

=> s = 2700 m

=> s = 2.7 km (Approximately)