Question

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$\huge \tt \color{red}Question:$

A bus starting from rest moves with a uniform acceleration of 0.1 m s‐² for 2 minutes.
Find
(a) the speed acquired,
(b) the distance travelled.

​

Answer 1

Question :-

A bus starting from rest moves with a uniform acceleration of 0.1 m/s² for 2 minutes.

Find

1. the speed acquired
2. the distance travelled

ANSWER

Given :-

A bus starts from rest moves with a uniform acceleration of 0.1 m/s² for 2 minutes.

Required to find :-

1. the speed acquired (final velocity {v})
2. the distance travelled (s)

Equation of motion :-

• v = u + at
• s = ut + ½ at²

Here,

v = final velocity

u = initial velocity

a = acceleration

s = displacement

t = time taken

Solution :-

Given that;

Acceleration of the bus = 0.1 m/s²

Time = 2 mins

convert 2 mins into seconds as the unit of time is seconds.

So,

Time = 2 mins

= 2 x 60 (1 min = 60 sec)

= 120 seconds

=> Time = 120 seconds

Since, the bus is starting from rest.

Initial velocity (u) = 0 m/s

Now,

→ v = u + at

v = at

v = (0.1)(120)

v = (1 x 120)/(10)

v = 12 m/s

So,

• Final velocity (v) = 12 m/s

Let's find the distance travelled !

Using the equation;

→ s = ut + ½at²

s = ½ at²

s = ½ (0.1) (120)²

s = ½ (1/10) (120)(120)

s = ½ (1 x 120 x 120)/(10)

s = (1 x 120 x 120)/(20)

s = (6 x 120)

s = 720 meters

Hence,

• Distance travelled = 720 meters

Answer 2

Answer:

Given :-

• A bus starting from rest moves with a uniform acceleration of 0.1 m/s² for 2 minutes.

To Find :-

• What is the speed acquired.
• What is the distance travelled.

Formula Used :-

$\clubsuit$ First Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

$\clubsuit$ Second Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2}at^2}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken
• s = Distance Covered

Solution :-

First, we have to convert time minutes into seconds :

$\implies \sf Time =\: 2\: minutes$

$\implies \sf Time =\: 2 \times 60\: seconds\: \: \bigg\lgroup \sf\bold{\pink{1\: minutes =\: 60\: seconds}}\bigg\rgroup$

$\implies \sf \bold{\purple{Time =\: 120\: seconds}}$

Now, we have to find the final velocity or speed of a bus :

Given :

• Initial Velocity = 0 m/s
• Acceleration = 0.1 m/s²
• Time Taken = 120 seconds

According to the question by using the formula we get,

$\longrightarrow \bf{v =\: u + at}$

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

$\longrightarrow \sf v =\: 0 + 0.1 \times 120$

$\longrightarrow \sf v =\: 0 + \dfrac{1}{10} \times 120$

$\longrightarrow \sf v =\: 0 + \dfrac{12\cancel{0}}{1\cancel{0}}$

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

$\longrightarrow \sf\bold{\red{v =\: 12\: m/s}}$

${\small{\bold{\underline{\therefore\: The\: speed\: acquired\: by\: a\: bus\: is\: 12\: m/s\: .}}}}$

Now, we have to find the distance travelled by a bus :

Given :

• Initial Velocity = 0 m/s
• Time Taken = 120 seconds
• Acceleration = 0.1 m/s²

According to the question by using the formula we get,

$\longrightarrow \bf s =\: ut + \dfrac{1}{2}at^2$

$\longrightarrow \sf s =\: 0(120) + \dfrac{1}{2} × 0.1(120)^2$

$\longrightarrow \sf s = 0 \times 120 + \dfrac{1}{2} × 0.1 \times 120 \times 120$

$\longrightarrow \sf s =\: 0 + \dfrac{1}{2} \times 0.1 \times 14400$

$\longrightarrow \sf s =\: 0 + \dfrac{1}{2} \times \dfrac{1}{10} \times 14400$

$\longrightarrow \sf s =\: 0 + \dfrac{1}{2} \times \dfrac{1440\cancel{0}}{1\cancel{0}}$

$\longrightarrow \sf s =\: 0 + \dfrac{1}{\cancel{2}} \times {\cancel{1440}}$

$\longrightarrow \sf s =\: 0 + 720$

$\longrightarrow \sf\bold{\red{ s =\: 720\: m}}$

${\small{\bold{\underline{\therefore\: The\: distance\: travelled\: by\: a\: bus\: is\: 720\: m\:.}}}}$