Physics, asked by Anonymous, 1 month ago

A train starting from rest attains a velocity of 54 km per hour in 5 minutes assuming that the acceleration is uniform find the acceleration and the distance travelled by the train for attaining this velocity​

Answers

Answered by rsagnik437
106

Answer :-

→ Acceleration of the train = 0.05 m/

→ Distance travelled = 2250 m

Explanation :-

We have :-

• Initial velocity (u) = 0 m/s

• Final velocity (v) = 54 km/h

Time taken (t) = 5 min = 300 sec

______________________________

Firstly, let's convert the unit of final velocity of the train (v) from "km/h" to "m/s" .

⇒ 1 km/h = 5/18 m/s

⇒ 54 km/h = 54(5/18)

⇒ 15 m/s

______________________________

By putting values in the 1st equation of motion, we would obtain acceleration of the train.

v = u + at

⇒ 15 = 0 + a(300)

⇒ 15 = 300a

⇒ a = 15/300

a = 0.05 m/s²

Now, we will calculate distance travelled by the train for attaining the final velocity by using the 3rd equation of motion .

- = 2as

⇒ (15)² - 0 = 2(0.05)s

⇒ 225 = 0.1s

⇒ s = 225/0.1

s = 2250 m

Answered by Anonymous
112

Provided that:

  • Initial velocity = 0 mps
  • Final velocity = 54 kmph
  • Time = 5 minutes

To calculate:

  • The acceleration
  • Distance

Solution:

  • The acceleration = 0.5 mps²
  • Distance = 2250 m

Using concepts:

  • Formula to convert kmph-mps
  • Formula to convert min-sec
  • Acceleration formula
  • Third equation of motion

Using formulas:

Formula to convert kmph-mps:

  • {\small{\underline{\boxed{\pmb{\sf{1 \: kmph \: = \dfrac{5}{18} \: mps}}}}}}

Formula to convert min-sec:

  • {\small{\underline{\boxed{\pmb{\sf{1 \: min \: = 60 \: sec}}}}}}

Acceleration formula:

  • {\small{\underline{\boxed{\pmb{\sf{a \: = \dfrac{v-u}{t}}}}}}}

  • We can use first equation of motion too for find acceleration.

First equation of motion:

  • {\small{\underline{\boxed{\pmb{\sf{v \: = u \: + at}}}}}}

Third equation of motion:

  • {\small{\underline{\boxed{\pmb{\sf{v^2 \: - u^2 \: = 2as}}}}}}

  • We can use second equation of motion too for calculating the distance travelled.

Second equation of motion:

  • {\small{\underline{\boxed{\pmb{\sf{s \: = ut + \dfrac{1}{2} \: at^2}}}}}}

Where, v denotes final velocity, u denotes initial velocity, a denotes acceleration, t denotes time taken, s denotes displacement or distance or height.

Required solution:

✡️ Firstly let us convert kmph-mps!

:\implies \sf 1 \: kmph \: = \dfrac{5}{18} \: mps \\ \\ :\implies \sf 54 \: kmph \: = 54 \times \dfrac{5}{18} \: mps \\ \\ :\implies \sf 54 \: kmph \: = 3 \times 5 \\ \\ :\implies \sf 54 \: kmph \: = 15 \: mps \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}

✡️ Now let us convert min to sec!

:\implies \sf 1 \: min \: = 60 \: sec \\ \\ :\implies \sf 5 \: min \: = 5 \times 60 \: sec \\ \\ :\implies \sf 5 \: min \: = 300 \: seconds \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}

✡️ Now let us find out the acceleration!

:\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{15-0}{300} \\ \\ :\implies \sf a \: = \dfrac{15}{300} \\ \\ :\implies \sf a \: = \dfrac{1}{20} \: ms^{-2} \\ \\ :\implies \sf a \: = 0.05 \: ms^{-2}

  • We can find acceleration by using first equation of motion too!

:\implies \sf v \: = u + \: at \\ \\ :\implies \sf 15 = 0 + a(300) \\ \\ :\implies \sf 15 = 300a \\ \\ :\implies \sf a \: = \dfrac{15}{300} \\ \\ :\implies \sf a \: = \dfrac{1}{20} \: ms^{-2} \\ \\ :\implies \sf a \: = 0.05 \: ms^{-2}

✡️ Now let us find out the distance!

:\implies \sf v^2 \: - u^2 \: = 2as \\ \\ :\implies \sf (15)^{2} - (0)^{2} = 2(0.05)(s) \\ \\ :\implies \sf 225 = 0.1s \\ \\ :\implies \sf 225/0.1 = s \\ \\ :\implies \sf 2250 \: m \: = s

  • We can use second equation of motion too for finding the distance travelled!

:\implies \sf s \: = ut \: + \dfrac{1}{2} \: at^2 \\ \\ :\implies \sf s \: = 0(300) + \dfrac{1}{2} \times 0.05(300)^{2} \\ \\ :\implies \sf s \: = 0 + \dfrac{1}{2} \times 0.05(90000) \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 0.05(90000) \\ \\ :\implies \sf s \: = 1 \times 0.05 \times 45000 \\ \\ :\implies \sf s \: = 2250 \: m

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