Question

a train acceleration is going uniformly from 36km/hr to 76km/hr in 10 seconds. It's acceleration is
A. 1m/sec² B. 10/msec² C. 5m/sec² D. 2•5m/sec²​

Answer 1

Answer:

Thus the acceleration is 1.12 m/s2.

Explanation:

The question should be that the train's velocity changes from 36 to 76 kmph.

Thus converting the speed to m/s we get 36×5/18= 10m/s which is v1

For 76kmph we get 76×5/18 = 21.12 m/s which is v2

Thus acceleration is v2-v1/T

So the acceleration is 21.2-10/10= 1.12 m/s2.

Answer 2

Provided that:

• Initial velocity = 36 kmph
• Final velocity = 76 kmph
• Time = 10 seconds

To determine:

• Acceleration

Solution:

• Acceleration = 1.11 ≈ 1.1 ≈ 1 m/s²

Using concepts:

• We can use either acceleration formula or first equation of motion to solve this question.

• Formula to convert kmph-mps.

Using formulas:

• Formula to convert kmph-mps:

• 1 kmph = 5/18 mps

• First equation of motion:

• v = u + at

• Acceleration formula:

• a = (v-u)/t

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

Required solution:

~ Firstly let us convert kmph-mps!

Converting initial velocity into metre per second firstly!

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

Now converting final velocity into metre per second!

$:\implies \sf 76 \times \dfrac{5}{18} \\ \\ :\implies \sf \dfrac{380}{18} \\ \\ :\implies \sf 21.11 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Therefore,

• Initial velocity = 10 mps
• Final velocity = 21.11 mps

~ Now let us find out the acceleration by using suitable formula!

By using first equation of motion...

$:\implies \sf v = u + at \\ \\ :\implies \sf 21.11 = 10 + a(10) \\ \\ :\implies \sf 21.11 - 10 = 10a \\ \\ :\implies \sf 11.11 = 10a \\ \\ :\implies \sf \dfrac{11.11}{10} \: = a \\ \\ :\implies \sf 1.111 \: = a \\ \\ :\implies \sf a \: = 1.111 \: ms^{-2}\\ \\ :\implies \sf Acceleration \: = 1.111 \: ms^{-2}$

By using acceleration formula...

$:\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{21.11-10}{10} \\ \\ :\implies \sf 11.11 = 10a \\ \\ :\implies \sf \dfrac{11.11}{10} \: = a \\ \\ :\implies \sf 1.111 \: = a \\ \\ :\implies \sf a \: = 1.111 \: ms^{-2}\\ \\ :\implies \sf Acceleration \: = 1.111 \: ms^{-2}$

• Therefore, acceleration = 1.11 mps sq. ≈ 1.1 mps sq. ≈ 1 mps sq. Henceforth, option (A.) is correct.

Knowledge booster:

About acceleration:

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