Question

A train accelarates Uniformly brom 36kmh-1 to 72kmh-1 in 10 seconds. Define acceleration, what do you mean by uniform motion​

Answer 1

Final Velocity = 72 km/hr

Time taken = 10 seconds

To find:

Distance travelled.

Conversion :

Convert unit of velocity from km/hr to

m/s.

1. 36 km/hr = 36 × 5/18 = 10 m/s

2. 72 km/hr = 72 × 5/18 = 20 m/s

Answer 2

Answer:

Question 1st: A train is uniformly accelerated from 36 kmph to 72 kmph in 10 seconds. What about its acceleration!?

Answer 1st:

Provided that:

• Initial velocity = 36 kmph
• Final velocity = 72 kmph
• Time = 10 seconds

To calculate:

• The acceleration

Solution:

• The acceleration = 1 m/s²

Using concepts:

• We can use either acceleration formula or first equation of motion to calculate the acceleration.

• Formula to convert kmph-mps.

Using formulas:

• Acceleration formula:

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

• First equation of motion:

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

• Formula to convert kmph-mps:

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

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

Required solution:

~ Firstly let us convert kmph-mps!

Converting 72 kmph-mps.

$:\implies \sf 72 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{72} \times \dfrac{5}{\cancel{{18}}} \: (Cancelling) \\ \\ :\implies \sf 4 \times 5 \\ \\ :\implies \sf 20 \: mps \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Converting 36 kmph-mps.

$:\implies \sf 36 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{36} \times \dfrac{5}{\cancel{{18}}} \: (Cancelling) \\ \\ :\implies \sf 2 \times 5 \\ \\ :\implies \sf 10 \: mps \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Therefore,

• Initial velocity = 10 mps
• Final velocity = 20 mps

~ Now let's find out the acceleration!

By using acceleration formula.

$:\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{20 - 10}{10} \\ \\ :\implies \sf a \: = \dfrac{10}{10} \\ \\ :\implies \sf a \: = 1 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 1 \: ms^{-2}$

By using first equation of motion.

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 20 = 10 + a(10) \\ \\ :\implies \sf 20 - 10 = 10a \\ \\ :\implies \sf 10 = 10a \\ \\ :\implies \sf a \: = \dfrac{10}{10} \\ \\ :\implies \sf a \: = 1 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 1 \: ms^{-2}$

Question 2nd: Define acceleration

Answer 2nd:

$\begin{gathered}\boxed{\begin{array}{c}\\ \bf What \: is \: acceleration? \\ \\ \sf The \: rate \: of \: change \: of \: velocity \: of \: an \\ \sf object \: with \: respect \: to \: time \\ \sf is \: known \: as \: acceleration. \\ \\ \sf \star \: Negative \: acceleration \: is \: known \: as \: deacceleration. \\ \sf \star \: Deceleration \: is \: known \: as \: retardation. \\ \sf \star \: It's \: SI \: unit \: is \: ms^{-2} \: or \: m/s^2 \\ \sf \star \: It \: may \: be \: \pm ve \: or \: 0 \: too \\ \sf \star \: It \: is \: a \: vector \: quantity \\ \\ \bf Conditions \: of \pm ve \: or \: 0 \: acceleration \\ \\ \sf (1) \: Positive \: acceleration: \: \sf When \: \bf{u} \: \sf is \: lower \: than \: \bf{v} \\ \sf (2) \: Negative \: acceleration: \: \sf When \: \bf{v} \: \sf is \: lower \: than \: \bf{u} \\ \sf (3) \: Zero \: acceleration: \: \sf When \: \bf{v} \: \sf and \: \bf{u} \: \sf are \: equal \end{array}}\end{gathered}$

Question 3rd: Define uniform motion.

Answer 3rd:

$\begin{gathered}\boxed{\begin{array}{c}\\ \bf What \: is \: motion? \\ \\ \sf Change \: in \: position \: over \: time \\ \sf is \: known \: as \: motion.\\ \sf There \: are \: different \: types \: of \\ \sf motion \: such \: as \: linear \: motion, \: uniform \: motion \\ \sf non \: uniform \: motion \: etc. \end{array}}\end{gathered}$

Uniform motion:

If a body covers equal distance in equal interval of time then it is known as uniform motion.

Non-uniform motion:

If a body covers unequal distance in equal interval of time then it is known as uniform motion.