Question

A bus starting from rest moves with
uniform acceleration of 0.1 m/s² for 2 min.
Find a) the speed acquired b)The distance
travelled​

Answer 1

Answer:

Kinematics

Explanation:

★ Given ;

» Uniform acceleration ( a ) = 0.1 m / s²

» Time ( t ) = 2 minutes = 120 seconds

» Initial speed ( U ) = 0

★ SPEED = ??

★ DISTANCE TRAVELLED = ??

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A ] Speed

From equation of motion

★ V = U + at

» V = 0 + 0.1 × 120

★ Speed ( V ) = 12 m / s

B ] Distance travelled

★ By second equation of motion ;

S = ut + 1 / 2 a t²

» S = 0 + 1 / 2 × 0.1 × ( 120 ) ²

» S = 1 / 2 × 1440

» S = 720 m

Distance traveled is 720 m .

Hope it helps you !

Answer 2

Given :

• Inital speed (u) = 0 m/sec

• Acceleration = 0.1m/s²

• Time = 2 min = 2× 60 = 120 seconds

To Find :

• The Speed Acquired

• The Distance Travelled

Solution :

Firstly, Let's find the speed acquired

From First Equation of Motion :

$\bigstar \: \: \large\boxed{ \sf \: v = u + at}$

$\longrightarrow \sf \: v = 0 + (0.1)(120)$

$\longrightarrow \sf \: v = 0 + 12$

$\longrightarrow \sf \red{ v = 12m {s}^{ - 1} }$

$\therefore$ Speed Acquired By Bus is 12 m/sec

Now, Let's find the Distance Travelled

From Third Equation of Motion :

$\bigstar \: \: \large\boxed{ \sf \: {v}^{2} = {u}^{2} + 2as}$

$\longrightarrow \sf \: ({12})^{2} = {0}^{2} + 2(0.1)(s)$

$\longrightarrow \sf \: 144 = 0 + 0.2s$

$\longrightarrow \sf \: - 0.2s = 0 - 144$

$\longrightarrow \sf \: - 0.2s = - 144$

$\longrightarrow \sf \: s = \dfrac{ - 144}{ - 0.2}$

$\longrightarrow \sf \: s = \dfrac{ 1440}{ 2}$

$\longrightarrow \sf \red{s = 720 \: metre}$

$\therefore$ Distance Travelled by Bus is 720 metres.