Question

calculate distance trvelled by bus which is moving at speed of 30m/s after traveling 10min bus is moving with 45m/s​

Answer 1

Answer :

➥ The distance travelled by bus = 25.87 km

Given :

➤ Intial velocity of a bus (u) = 30 m/s

➤ Final velocity of a bus (v) = 45 m/s

➤ Time taken by bus (t) = 10 min

To Find :

➤ Distance travelled by a bus (s) = ?

Solution :

◈ Time taken (t) = 10 min = 10 × 60 = 600 sec

To find distance travelled by a bus first we need to find the acceleration of a bus then after we find the distance travelled by a bus from second equation of motion.

Acceleration of a bus

Acceleration is given by

$\tt{:\implies a = \dfrac{v - u}{t} }$

$\tt{:\implies a = \dfrac{45 - 30}{600} }$

$\tt{:\implies a = \cancel{\dfrac{15}{600}} }$

$\bf{:\implies \underline{ \: \: \underline{ \green{ \: \: a = \dfrac{1}{40} \: m/s^2 \: \: }} \: \: }}$

Now ,

We find distance travelled by a bus

From second equation of motion

$\tt{:\implies s = ut + \dfrac{1}{2} a {t}^{2} }$

$\tt{:\implies s = 30 \times 600 \times \dfrac{1}{2} \times \dfrac{1}{40} \times 600 \times 600}$

$\tt{:\implies s = 18000 + \dfrac{1}{2} \times \dfrac{1}{40} \times 360000}$

$\tt{:\implies s = 18000 + \dfrac{1}{ \cancel { \: 2 \: }} \times \cancel{15750}}$

$\tt{:\implies s = 18000 + 1 \times 7875}$

$\tt{:\implies s =18000 + 7875}$

$\tt{:\implies s = 25875 \: m}$

Convert into km :

$\tt{:\implies s = \cancel{\dfrac{25875}{1000}} }$

$\bf{:\implies \underline{ \: \: \underline{ \purple{ \: \: s = 25.87 \: km \: \: }} \: \: }}$

Hence, the distance travelled by a bus is 25.87 km.

⠀

Some releted equations :

⪼ s = ut + ½ at²

⪼ v = u + at

⪼ v² = u² + 2as

Answer 2

$\large{\underline{\underline{\red{\sf{\hookrightarrow Given :-}}}}}$

• A bus has a Initial velocity of 30m/s .
• It has a final Velocity of 45m/s.

$\large{\underline{\underline{\red{\sf{\hookrightarrow To\:Find: }}}}}$

• The distance travelled by bus in 10minutes.

$\large{\underline{\underline{\red{\sf{\hookrightarrow Formula\:Used: }}}}}$

We will use second and firsrt equⁿs of motion :

$\large\pink{\underline{\boxed{\purple{\tt{\dag \:v\:\:=\:\:u\:\:+\:\:at }}}}}$

$\large\pink{\underline{\boxed{\purple{\tt{\dag \:s\:\:=\:\:ut\:\:+\:\:\dfrac{1}{2}at^2}}}}}$

$\large{\underline{\underline{\red{\sf{ \hookrightarrow Answer:}}}}}$

Here ,

• ❒$\tt Initial \: velocity\:=\:30m/s$
• ❒$\tt Final \: velocity \:=\:45m/s$
• ❒ $\tt Time\:taken\:=\:10min.$

Now , convert time in SI unit , 10min = 10 × 60s = 600s.

$\tt Put \: these\:values\:in\:first\:equ^n\:of\:motion$

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

$\tt :\implies 45m/s=30m/s + a\times 600s$

$\tt :\implies 600a = 45m/s-30m/s$

$\tt :\implies 600a = 15m/s$

$\tt :\implies a =\dfrac{\cancel{15}}{\cancel{600}}m/s^2$

$\underline{\boxed{\red{\tt{\longmapsto a\:\:=\:\:\dfrac{1}{40}ms^{-2}}}}}$

$\green{\bf \leadsto Hence\:accl^n\:is\:\dfrac{1}{40}ms^{-2}}$

_________________________________________

Now , things we have now is ,

• ❒Acclⁿ = 1/40m/s².
• ❒ Initial velocity = 30m/s.
• ❒ time = 600s .

$\tt Now \:use \:2nd\: equ^n\: of\: motion ,$

$\tt:\implies s=ut+\dfrac{1}{2}at^2$

$\tt:\implies s=30m/s\times600s+\dfrac{1}{2}\times \dfrac{1}{40}ms^{-2}\times(600s)^2$

$\tt:\implies s = 18000+\dfrac{1}{\cancel{2}}\times \cancel{600s}^{150}\times\cancel{60}\cancel{0}s\times\dfrac{1}{\cancel{4}\cancel{0}}ms^{-2}$

$\tt:\implies s= 18000 m + 7875m$

$\tt:\implies s = 25785m$

$\underline{\boxed{\red{\tt{\longmapsto Distance\:\:=\:\:25.875km}}}}$

$\orange{\boxed{\green{\bf{\dag\:Hence\:the\: distance\:covered\:is\:25.875\:km.}}}}$