Question

if a train travelling at 10 m/s starts to accelerate at 1 m/s² for 15 seconds on a straight track calculate its final speed in m/s

Answer 1

Provided that:

• Initial velocity = 10 mps
• Acceleration = 1 mps sq.
• Time = 15 seconds

To calculate:

• Final velocity

Solution:

• Final velocity = 25 mps

Using concept(s):

• To solve this question we can use either first equation of motion or acceleration formula.

Using formula(s):

• Acceleration formula:

• a = (v-u)/t

• First equation of motion:

• v = u + at

Where, a denotes acceleration, v denotes final velocity, u denotes initial velocity, t denotes time taken.

Required solution:

~ By using acceleration formula!

»»» a = (v-u)/t

»»» 1 = (v-10)/15

»»» 1 × 15 = v - 10

»»» 15 = v - 10

»»» 15 + 10 = v

»»» 25 = v

»»» v = 25 mps

»»» Final velocity = 25 mps

~ By using first equation of motion!

»»» v = u + at

»»» v = 10 + 1(15)

»»» v = 10 + 15

»»» v = 25 mps

»»» Final velocity = 25 mps

Additional information:

$\begin{gathered}\boxed{\begin{array}{c}\\ \bf What \: is \: acceleration? \\ \\ \sf The \: rate \: of \: change \: of \: velocity \: of \: an \\ \sf object \: with \: respect \: to \: time \\ \sf is \: known \: as \: acceleration. \\ \\ \sf \star \: Negative \: acceleration \: is \: known \: as \: deceleration. \\ \sf \star \: Deceleration \: is \: known \: as \: retardation. \\ \sf \star \: It's \: SI \: unit \: is \: ms^{-2} \: or \: m/s^2 \\ \sf \star \: It \: may \: be \: \pm ve \: or \: 0 \: too \\ \sf \star \: It \: is \: a \: vector \: quantity \\ \\ \bf Conditions \: of \pm ve \: or \: 0 \: acceleration \\ \\ \sf \odot \: Positive \: acceleration: \: \sf When \: \bf{u} \: \sf is \: lower \: than \: \bf{v} \\ \sf \odot \: Negative \: acceleration: \: \sf When \: \bf{v} \: \sf is \: lower \: than \: \bf{u} \\ \sf \odot \: Zero \: acceleration: \: \sf When \: \bf{v} \: \sf and \: \bf{u} \: \sf are \: equal \end{array}}\end{gathered}$

Answer 2

Question :-

• A train travelling 10 m/s starts to accelerate at 1 m/s² for 15 seconds on a straight track. Calculate its final speed in m/s .

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Answer :-

• Final Velocity = 25 m/s .

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Explanation :-

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Given :-

• Initial Velocity = 10 m/s
• Acceleration = 1 m/s sq.
• Time = 15 seconds

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To Find :-

• Final Velocity = ?

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Solution :-

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∓ Here, we have to calculate the Final Velocity of the train .

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★ Formula :- ᴀ = ( ᴠ - ᴜ ) / ᴛ

• A denotes to Acceleration .
• V denotes to Final Velocity .
• U denotes to Initail Velocity .
• T denotes to Time .

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∓ Now, let's substitute ------

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$\longrightarrow \: \bf{ᴀ = ( \: v - u \: ) \: / \: ᴛ} \\ \\ \longrightarrow \: \tt{1 = ( \: v - 10 \: ) \: / \: 15} \\ \\ \longrightarrow \: \tt{1 \times 15 = v - 10} \\ \\ \longrightarrow\: \tt{15 = v - 10} \\ \\ \longrightarrow\: \tt{v = 15 + 10} \\ \\ \longrightarrow\: \sf \red{v = 25 \: m /s}$

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Hence :-

• Final Velocity = 25 m/s .

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