Question

A bus starting from rest attains a speed of 108 km/hr in 10 sec . calculate the ac produced & distance travelled

Answer 1

Given :

➳ Initial speed = zero (i.e., rest)

➳ Final speed = 108kmph

➳ Time of journey = 10s

To Find :

• Acceleration
• Distance travelled

Solution :

➛ Acceleration is defined as the rate of change in speed.

➛ It is a vector quantity having both magnitude as well as direction.

➛ SI unit : m/s²

➔ 1 km/h = 5/18 m/s

➔ 108 km/h = 108 × 5/18 = 30 m/s

✭ Acceleration of bus :

⇒ $\tt{v=u+at}$

⇒ $\tt{30=0+a(10)}$

⇒ $\tt{a=\dfrac{30}{10}}$

⇒ $\underline{\boxed{\bf{a=3\:ms^{-2}}}}$

✭ Distance travelled by bus :

⇒ $\tt{v^2-u^2=2as}$

⇒ $\tt{(30)^2-(0)^2=2(3)s}$

⇒ $\tt{900=6s}$

⇒ $\underline{\boxed{\bf{s=150\:m}}}$

Answer 2

$\color{red}{\large\underline{\underline\mathtt{Question:}}}$

A bus starting from rest attains a speed of 108 km/hr in 10 sec .calculatethe ac produced & distance travelled.

______________________________________

$\color{purple}{\large\underline{\underline\mathtt{To\:Find:}}}$

• $\mathtt{The\:distance\:covered}$
• $\mathtt{The\:acceleration\:produced}$

______________________________________

$\color{blue}{\large\underline{\underline\mathtt{Concept:}}}$

$\textit{Here ,the first we have to change the velocity}$ $\textit{from km/hr to m/s.}$

______________________________________

$\color{blue}{\large\underline{\underline\mathtt{Given:}}}$

• $\mathrm{Initial\:velocity \rightarrow 0m\:s^{-1}}$
• $\mathrm{Final\:velocity \rightarrow 108km\:hr^{-1} = 30m\:s^{-1}}$

______________________________________

$\color{magenta}{\large\underline{\underline\mathtt{Solution:}}}$

$\underline{Distance\:covered}$

From the relation ,

$v^{2} = u^{2} + 2as$

Putting the value in the formula , we get:

$\Rightarrow 30^{2} = 0^{2} + 2 \times 3s$

$\Rightarrow 900 = 6s$

$\Rightarrow \dfrac{900}{6} = s$

$\Rightarrow \dfrac{\cancel{900}}{\cancel{6}} = s$

$\Rightarrow 150 m = s$

______________________________________

$\underline{Acceleration}$

We know ,

$v = u + at$

putting the value in the formula, we get :

$\Rightarrow 30 = 0 + a \times 10$

$\Rightarrow 30 = 10a$

$\Rightarrow \dfrac{30}{10} = a$

$\Rightarrow \dfrac{\cancel{30}}{\cancel{10}} = a$

$\Rightarrow 3 m\:s{-2} = a$

______________________________________

• $\mathtt{The\:distance\:covered = 300m}$
• $\mathtt{The\:acceleration\:produced = 3m\:s^{-2}}$