Question

A bus starting from rest moves with uniform acceleration of 0.1m/s2 for 2 minutes.Find the speed acquired and distance travelled.

Answer 1

ProvidEd that:

• Initial velocity = 0 m/s
• Acceleration = 0.1 m/s sq.
• Time = 2 minutes

To calculaTe:

• Speed acquired
• Distance travelled

SoluTion:

• Speed acquired = 12 m/s
• Distance travelled = 720 m

KnowlEdge required:

• SI unit of speed = m/s
• SI unit of time = seconds
• SI unit of distance = m
• SI unit of acceleration = m/s²

UsiNg concepts:

• Formula to convert min-sec.

• First equation of motion to calculate the speed acquired.

• Third equation of motion / Second equation of motion to calculate distance.

Using formulas:

• 1 min = 60 seconds

• 1st eqⁿ of motion → v = u + at

→ 2nd eqⁿ of motion → s = ut + ½ at²

→ 3rd eqⁿ of motion → v² - u² = 2as

Required solution:

~ Firstly let us convert min-sec!

⇝ 1 min = 60 seconds

⇝ 2 minutes = 2 · 60

⇝ 2 minutes = 120 seconds

• Henceforth, converted!

~ Now by using first equation of motion let us find out the speed acquired!

⇝ v = u + at

⇝ v = 0 + 0.1(120)

⇝ v = 0 + 12

⇝ v = 12 m/s

⇝ Final velocity = 12 m/s

• Speed acquired = 12 m/s

~ Now finding distance travelled!

By using second equation of motion...

⇝ s = ut + ½ at²

⇝ s = 0(120) + ½ · 0.1(120)²

⇝ s = 0 + ½ · 0.1 · 14400

⇝ s = 0 + ½ · 1440

⇝ s = 0 + 720

⇝ s = 720 metres

⇝ Distance = 720 metres

• Distance travelled = 720 m

By using third equation of motion...

⇝ v² - u² = 2as

⇝ (12)² - (0)² = 2(0.1)(s)

⇝ 144 - 0 = 0.2s

⇝ 144 = 0.2s

⇝ 144/0.2 = s

⇝ 1440/2 = s

⇝ 720 = s

⇝ s = 720 metres

⇝ Distance = 720 metres

• Distance travelled = 720 m

AdditioNal information:

Difference between speed and velocity is mentioned below:

$\begin{gathered}\boxed{\begin{array}{c|cc}\bf Speed&\bf Velocity\\\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}&\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}\\\sf The \: distance \: travelled \: by &\sf The \: distance \: travelled \: by \\ \sf \: a \: body \: per \: unit \: time&\sf \: a \: body \: per \: unit \: time \\ &\sf in \: a \: given \: direction \\\\\sf It \: is \: scalar \: quantity. &\sf It \: is \: vector \: quantity \\\\\sf It \: is \: positive \: always &\sf It \: can \: be \: \pm \: \& \: 0 \: too \\\\\sf Speed \: = \dfrac{Distance}{Time} &\sf Velocity \: = \dfrac{Displacement}{Time} \end{array}}\end{gathered}$

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Hello dear users, with due respect I want to say that we can find out the distance by using either third equation of motion or second equation of motion, choice may vary! I solve this part by both methods for you, you can use any one of them! Thank you!