Question

5) A bus starts from rest and moving with uniform acceleration attains a velocity of 50 km/h in 3 minutes. Find the acceleration and the distance travelled.​

Answer 1

$\large \bf \clubs \:  Given :-$

A bus starts from rest and moving with uniform acceleration attains a velocity of 50 km/h in 3 minutes.

• Initial Velocity = u = 0 m/s

• Final Velocity = v = 50 km/h

• Time = t = 3 mins

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$\large \bf \clubs \:  To \: Find :-$

• Acceleration of the Bus

• Distance Traveled By Bus

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$\large \bf \clubs \:  Important \: Conversions :-$

✏ Converting Final Velocity in m/s :

We Have,

$\text{Final Velocity = v = 50 km/h} \\ \\ = \bigg( 50 \times \dfrac{5}{18} \bigg) \text{m/s} \\ \\ = \bigg( \dfrac{250}{18} \bigg) \text{m/s}$

$:\longmapsto \text{v = 13.89 m/s}$

✏ Converting Time in second :

We Have,

$\text{Time = t = 3 mins } \\ \\ = (3 \times 60) \text{ second}$

$:\longmapsto \text{t = 180 s}$

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$\large \bf \clubs \:  Solution :-$

✏ Calculating Acceleration :

$\orange{ \bf Applying : v = u + at}$

$:\longmapsto 13.89 = 0 + \text{a }\times 180$

$:\longmapsto \text{13.89 = 180a}$

$:\longmapsto \text a = \dfrac{13.89}{180}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf a = 0.077 \: m/ {s}^{2} } }}}$

✏ Calculating Distance Traveled :

Let Distance Traveled = s meter

$\orange{ \bf Applying : v^2-u^2 = 2as}$

$:\longmapsto 13.89^2-0^2=2(0.077)(\text s)$

$:\longmapsto 192.93-0=0.154 \text s$

$\text s = \dfrac{192.93}{0.154}$

$\large:\longmapsto \text{ s = 1252.79m}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf s = 1.25 \: km(approx)} }}}$

Hence ,

$\pink{\begin{cases}\bf Acceleration = 0.077 \: m/s {}^{2} \\ \\\bf Distance \: Traveled = 1.25 \:km\end{cases}}$

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$\large \bf \clubs \: Additional\: Info :-$

✏ Three Equations of Motion :

$\large \qquad \boxed{\boxed{\begin{array}{cc} \maltese \: \: \bf v = u + at \\ \\ \maltese \: \: \bf s = ut + \dfrac{1}{2}a {t}^{2} \\ \\ \maltese \: \: \bf{v}^{2} - {u}^{2} = 2as\end{array}}}$

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Answer 2

Answer :

$\:$

• Acceleration of a bus is 0.077 m/s².

• Distance travelled by a bus is 1247.4 m.

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Explanation :

$\:$

Given :

$\:$

A bus starts from rest and moving with

uniform acceleration attains a velocity of 50 km/h in 3 minutes.

$\:$

To Find :

$\:$

• Acceleration of a bus and distance travelled by a bus?

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Solution :

$\:$

• Here, we have initial velocity of a bus (u) = 0 m/s, final velocity of a bus (v) = 50 km/h = (50 × 1/18) m/s = 13.88 m/s and time taken by a bus (t) = 3 minutes = (3 × 60) seconds = 180 seconds. We have to find out acceleration of a bus (a) and distance travelled by a bus (s)? So, here we will use first and third equation of motion to find acceleration and distance travelled. So, let's solve it!

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Equations Used :

$\:$

• $\underline{\boxed{\bf{\pink{\pmb{v = u + at}}}}}$

• $\underline{\boxed{\bf{\blue{\pmb{s = ut + \dfrac{1}{2}at^2}}}}}$

$\:$

Where,

$\:$

• v denotes final velocity.

• u denotes initial velocity.

• a denotes acceleration.

• t denotes time taken.

• s denotes distance travelled.

$\:$

$\footnotesize\underline{\bigstar{\bm{\pmb{\:Putting\:all\:values\:in\:1st\:equation\:of\:motion\::}}}}$

$\\ :\implies \:\sf 13.88 = 0 + a\big(180\big)$

$\\ :\implies \:\sf 13.88 = 180a$

$\\ :\implies \:\sf a = {\cancel{\dfrac{13.88}{180}}}$

$\\ :\implies \:\underline{\boxed{\frak{\pmb{\red{a = 0.077}}}}}\:\green{\bigstar}$

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• Therefore, acceleration of a bus (a) is 0.077 m/s².

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Now,

$\:$

$\footnotesize\underline{\bigstar{\bm{\pmb{\:Putting\:all\:values\:in\:2nd\:equation\:of\:motion\::}}}}$

$\\ :\implies \:\sf s = 0\big(180\big) + \dfrac{1}{2}\:\times\:\big(0.077\big)\big(180\big)^2$

$\\ :\implies \:\sf s = 0 + \dfrac{1}{2}\:\times\:0.077\:\times\:32400$

$\\ :\implies \:\sf s = \dfrac{1}{\cancel{2}}\:\times\:0.077\:\times\:\cancel{32400}$

$\\ :\implies \:\sf s = 0.077 \:\times\:16200$

$\\ :\implies \:\underline{\boxed{\frak{\pmb{\green{s = 1247.4}}}}}\:\red{\bigstar}$

$\:$

• Therefore, distance travelled by a bus (s) is 1247.4 m.

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