Question

plz help me..................... ​

Answer 1

Given:-

• A bus starts from rest.
• It achieves. a velocity of 12m/s.
• It has a acclⁿ of 4m/s².

To Find:-

• The distance travelled by the bus.

Formula Used:-

We will use third equⁿ of motion , which is ,

$\large\blue{\underline{\boxed{\red{\tt{\dag\:\:2a s\:\:=\:\:v^2\:\:-\:\:u^{2}\:\: }}}}}$

Where ,

• s is distance .
• t is time.
• a is acceleration.
• u is Initial velocity.

Calculation:-

Here given that ,

➤ $\pink{\bf Initial \:velocity\: of\: bus\: = 0m/s.}$

➤ $\purple{\bf Acceleration\: of\: bus\: = 4m/s².}$

➤ $\pink{\bf Final \:velocity \:of\: bus\: = 12m/s.}$

➤ $\purple{\bf Distance\: travelled\: by\: bus = ? }$

Now , putting the values in above formula stated ,

$\tt :\implies 2as=v^2-u^2$

$\tt :\implies 2\times4ms^{-2}\times s = (12ms^{-1}-(0)^2$

$\tt :\implies 8ms^{-2}\times s = 144m^2s^{-2}$

$\tt :\implies s =\dfrac{\cancel{144m^2s^{-2}}^8}{\cancel{8ms^{-2}}}$

$\underline{\boxed{\red{\tt{\longmapsto s=\:\:=\:\:18m\:\:}}}}$

$\green{\tt Hence\:the\:bus\: travelled\:18m.}$

Answer 2

Question:-

A bus starting form rest achieves a velocity of 12 m/s. The acceleration of the bus is 4 m/s². Then find the distance covered by the bus.

Formula Used:-

v² = u² + 2as

where

• v = final velocity
• u = initial velocity
• a = acceleration
• s = displacement

Answer:-

Given:

• u = 0m/s
• v = 12m/s
• a = 4m/s²

So applying the formula,

v² = u² + 2as

=> (12)² = (0)² + [ 2(4) * s]

=> 144 = 8 * s

=> s = 144/8 m

=> s = 18 m

Ans. Displacement = 18 m