Question

A motorbike running at 25 m/s is slowed down to 15 m/s by the application of brakes over
a distance of
40 m. If the brakes produce uniform retardation, then calculate:
(i) The retardation of the motorbike
(ii) Total distance travelled by the bike till it comes to rest.
(ii) Time which the bike comes to rest

Answer 1

Answer -

Retardation = 5 m/s^2

Total distance = 62.5 m

Total time taken to come to rest = 5 seconds

━━━━━━━━━━━━

Solution -

Given -

• v = 15 m/s
• u = 25 m/s
• s = 40 m

where

u is initial velocity.

v is final velocity.

s is distance travelled.

To find -

• Retardation
• Total Distance travelled till it came to rest
• Time taken to come to rest

━━━━━━━━━━━━

$\sf\purple{Calculating \:retardation- }$

• v = 15 m/s
• u = 25 m/s
• s = 40 m

Substituting the value in 3rd equation of motion

$\tt v^2 = u^2 + 2as$

$\implies$$\tt 15^2 = 25^2 + 2 \times a \times 40$

$\implies$$\tt 225 = 625 + 2 \times a \times 40$

$\implies$$\tt -400 = 2 \times a \times 40$

$\implies$$\tt 80a = -400$

$\implies$$\tt a = \dfrac{-400}{80}$

$\implies$$\tt a = - 5 m/s^2$

Retardation of motor bike is 5 m/s^2

━━━━━━━━━━━━

$\sf\purple{Calculating \:total \:distance \:travelled}$

• u = 25 m/s
• v = 0 m/s ( because we have to calculate distance travelled when it came to rest )
• a = -5 m/s^2

Substituting the value in 3rd equation of motion -

$\tt v^2 = u^2 + 2as$

$\implies$$\tt 0 = 25^2 + 2 \times (-5) \times s$

$\implies$$\tt 625 = 10s$

$\implies$$\tt s = 62.5 m$

━━━━━━━━━━━━

$\sf\purple{Calculating\: Time\: taken\: to\: comes \:to\: rest -}$

• u = 25 m/s
• v = 0 m/s ( because we have to calculate time taken to came to rest )
• a = -5 m/s^2

Substituting the value in 1st equation of motion -

$\tt v = u + at$

$\implies$$\tt 0 = 25 + (-5)t$

$\implies$$\tt 25 = 5t$

$\implies$$\tt t = 5 sec$

━━━━━━━━━━━━

Answer 2

$\huge\sf { \dag Question }$

A motorbike running at 25 m/s is slowed down to 15 m/s by the application of brakes over a distance of 40 m. If the brakes produce uniform retardation, then calculate:

• (i) The retardation of the motorbike

• (ii) Total distance travelled by the bike till it comes to rest.

• (iii) Time which the bike comes to rest

$\huge\sf { \dag Solution }$

${\underline {\underline {\rm {\red { Given \: information: }}}}}$

• $\sf\pink { u ( initial \: velocity ) = 25m/s }$

• $\sf\pink { v ( final \: velocity ) = 15m/s}$

• $\sf\pink { s ( distance \: travelled ) = 40m}$

___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___

☆ The retardation of the motorbike

According to the third equation of motion -

${ \qquad {\underline {\underline {\boxed {\rm {\purple { 2as = {v}^{2} - {u}^{2} }}}}}}}$

Where,

• $\sf { s ( distance ) = 40m}$

• $\sf { v ( final \: velocity ) = 15m/s}$

• $\sf { u ( initial \: velocity ) = 25m/s }$

• $\sf { a ( acceleration ) =?}$

Substituting values to find retardation-

$\sf\red { \qquad 2as ={v}^{2} - {u}^{2} }$

$\sf { \qquad \leadsto 2 \times a \times 40 ={15}^{2} - {25}^{2} }$

$\sf { \qquad \leadsto 80 \times a = 225 - 625}$

$\sf { \qquad \leadsto 80a = -400}$

$\sf { \qquad \leadsto a = \cancel {\dfrac{-400}{80}} = -5m/{s}^{2} }$

Therefoere, retardation of the motorbike = $\sf { 5m/{s}^{2} }$

_______________________________

☆ Total distance travelled by the bike till it comes to rest.

According to the third equation of motion -

${ \qquad {\underline {\underline {\boxed {\rm {\purple { 2as = {v}^{2} - {u}^{2} }}}}}}}$

Where,

• $\sf { s ( total \: distance ) = ? }$

• $\sf { v ( final \: velocity ) = 0m/s}$

[ Here, v ( final velocity ) will be zero because the question is asking total distance travelled by the bike till it comes to rest. ]

• $\sf { u ( initial \: velocity ) = 25m/s }$

• $\sf{ a ( acceleration ) = -5m/{s}^{2}}$

Substituting values to find total distance travelled-

$\sf\red { \qquad 2as ={v}^{2} - {u}^{2} }$

$\sf { \qquad \leadsto 2 \times -5 \times s ={0}^{2} - {25}^{2} }$

$\sf { \qquad \leadsto -10 \times s = 0 - 625 }$

$\sf { \qquad \leadsto -10 \times s = -625 }$

$\sf { \qquad \leadsto -10s = -625 }$

$\sf { \qquad \leadsto s = \cancel { \dfrac{-625}{-10}} = 62.5m }$

Therefore, total distance travelled by the bike till it comes to rest is 62.5m.

_______________________________

☆ Time which the bike comes to rest

According to the first equation of motion -

${ \qquad {\underline {\underline {\boxed {\rm {\purple { v = u + at }}}}}}}$

Where,

• $\sf { v ( final \: velocity ) = 0m/s}$

[ Here, v ( final velocity ) will be zero because the question is asking time which the bike comes to rest . ]

• $\sf { u ( initial \: velocity ) = 25m/s }$

• $\sf{ a ( acceleration ) = -5m/{s}^{2}}$

• $\sf{ t (time ) = ? }$

Substituting values to find time which the bike comes to rest-

$\sf\red { \qquad v = u + at }$

$\sf { \qquad \leadsto 0 = 25 + ( -5 \times t ) }$

$\sf { \qquad \leadsto 0 = 25 + ( -5 t ) }$

$\sf { \qquad \leadsto 0 = 25 + ( -5 t )}$

$\sf { \qquad \leadsto 5t = 25 }$

$\sf { \qquad \leadsto t = \cancel { \dfrac{25}{5}} = 5seconds }$

_______________________________

$\huge\sf { \dag Required \: Answers: }$

• The retardation of the motorbike = $\sf { 5m/{s}^{2} }$

• Total distance travelled by the bike till it comes to rest = $\sf { 62.5m }$

• Time which the bike comes to rest = $\sf { 5 seconds }$

______________________________