Question

a train accelerates uniforlmy fron 36 km/h to 72 km/h in 20s. Find the distance travelled by it​

Answer 1

Provided that:

• Initial velocity = 36 km/h
• Initial velocity = 72 km/h
• Time taken = 20 seconds

To calculate:

• The distance

Solution:

• The distance = 300 m

Using concept:

• Acceleration formula
• Third equation of motion

Or can also use

• Second equation of motion
• Formula to convert kmph

Using formula:

• Acceleration is given by,

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

• Second equation of motion is given by,

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

• Third equation of motion is given by,

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

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

Required solution:

~ Firstly let us convert initial velocity into metre per second!

$:\implies \sf 36 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{36} \times \dfrac{5}{\cancel{{18}}} \\ \\ :\implies \sf 2 \times 5 \\ \\ :\implies \sf 10 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

~ Now let us convert final velocity into metre per second!

$:\implies \sf 72 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{72} \times \dfrac{5}{\cancel{{18}}} \\ \\ :\implies \sf 4 \times 5 \\ \\ :\implies \sf 20 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Therefore,

• Initial velocity is 10 m/s
• Final velocity is 20 m/s

~ Now let's calculate acceleration!

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Change \: in \: time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{20-10}{20} \\ \\ :\implies \sf a \: = \dfrac{10}{20} \\ \\ :\implies \sf a \: = \cancel{\dfrac{10}{20}} \\ \\ :\implies \sf a \: = 0.5 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 0.5 \: ms^{-2}$

Therefore, acceleration is 0.5 m/s sq.

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~ Now let us calculate the distance by using second equation of motion!

$:\implies \sf s \: = ut \: + \dfrac{1}{2} at^2 \\ \\ :\implies \sf s \: = (10)(20) + \dfrac{1}{2} \times (0.5)(20)^{2} \\ \\ :\implies \sf s \: = (10)(20) + \dfrac{1}{2} \times 0.5 \times 20 \times 20 \\ \\ :\implies \sf s \: = (10)(20) + \dfrac{1}{2} \times 0.5 \times 400 \\ \\ :\implies \sf s \: = 200 + \dfrac{1}{2} \times 0.5 \times 400 \\ \\ :\implies \sf s \: = \cancel{200} + \dfrac{1}{\cancel{{2}}} \times 0.5 \times 400 \\ \\ :\implies \sf s \: = 100 + 1 \times 0.5 \times 400 \\ \\ :\implies \sf s \: = 100 + 1 \times 200 \\ \\ :\implies \sf s \: = 100 + 200 \\ \\ :\implies \sf s \: = 300 \: m \\ \\ :\implies \sf Distance \: = 300 \: metres$

~ Now let us calculate the distance by using third equation of motion!

$:\implies \sf 2as \: = v^2 \: - u^2 \\ \\ :\implies \sf 2(0.5)(s) \: = (20)^{2} - (10)^{2} \\ \\ :\implies \sf 2(0.5)(s) \: = 400 - 100 \\ \\ :\implies \sf 2(0.5)(s) \: = 300 \\ \\ :\implies \sf 1s \: = 300 \\ \\ :\implies \sf s \: = 300 \: m \\ \\ :\implies \sf Distance \: = 300 \: metres$

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• We can solve this question by using any one of the equation. I solved this question by both methods for you. That's your choice to use which equation.