Question

a bus starting from rest moves with a uniform acceleration of 0.1m/s square for 2 mins. find the speed acquired and total distance travelled​

Answer 1

Given :-

• A bus starts from rest .
• It has a accelⁿ of 0.1 m / s².
• It travels for 2 minutes .

To Find :-

• The speed accquired during this time .

Formula Used :-

• We will use first equation of motion , which is stated as ,

$\large\qquad\qquad\boxed{\red{\bf \green{\dag} v\:\:=\:\:u\:\:+\:\:at}}$

Where ,

• v is final velocity.
• u is Initial velocity.
• a is acceleration .
• t is time taken.

Calculation :-

Here , as per the given data ,

• u = 0m/s .
• t = 2min = 120 s .
• a = 0.1 m / s² .
• v = ? .

Now , put the value in above formula stated ,

$\tt:\implies v = u + at$

$\tt:\implies v = 0m/s + 0.1m/s^2 \times 120s$

$\tt:\implies v = \bigg\lgroup \dfrac{1}{1\cancel0}ms^{-2}\times 12\cancel0s\bigg\rgroup$

$\tt:\implies v = 0m /s +12m/s$

$\underline{\boxed{\red{\tt\longmapsto Final\: Velocity=12m/s}}}$

$\boxed{\green{\bf\pink{\dag}Hence\:the\:final\: velocity\:is\:12ms^{-1}}}$

Answer 2

Answer:

Given :-

A bus starts from rest .

It has a accelⁿ of 0.1 m / s².

It travels for 2 minutes .

To Find :-

The speed accquired during this time .

Formula Used :-

We will use first equation of motion , which is stated as ,

\large\qquad\qquad\boxed{\red{\bf \green{\dag} v\:\:=\:\:u\:\:+\:\:at}}

†v=u+at

Where ,

v is final velocity.

u is Initial velocity.

a is acceleration .

t is time taken.

Calculation :-

Here , as per the given data ,

u = 0m/s .

t = 2min = 120 s .

a = 0.1 m / s² .

v = ? .

Now , put the value in above formula stated ,

\tt:\implies v = u + at:⟹v=u+at

\tt:\implies v = 0m/s + 0.1m/s^2 \times 120s:⟹v=0m/s+0.1m/s

2

×120s

\tt:\implies v = \bigg\lgroup \dfrac{1}{1\cancel0}ms^{-2}\times 12\cancel0s\bigg\rgroup:⟹v=

⎩

⎪

⎪

⎪

⎧

1

0

1

ms

−2

×12

0

s

⎭

⎪

⎪

⎪

⎫

\tt:\implies v = 0m /s +12m/s:⟹v=0m/s+12m/s

\underline{\boxed{\red{\tt\longmapsto Final\: Velocity=12m/s}}}

⟼FinalVelocity=12m/s

\boxed{\green{\bf\pink{\dag}Hence\:the\:final\: velocity\:is\:12ms^{-1}}}

†Hencethefinalvelocityis12ms

−1