Question

A bike is moving at a speed of 30 km/h. Brakes are applied in order to
produce uniform retardation of 0.5 m/s. Find the distance covered before it stops.
​

Answer 1

Given :

▪ Initial speed of bike = 30kmph

▪ Retardation due to applied brakes = 0.5m/s²

To Find :

▪ Distance covered by bike before it stops.

Concept :

☞ This question is completely based on the concept of stopping distance.

☞ Stopping distance is defined as the distance covered by body before it is brought to rest.

☞ In this question, Acceleratiom has said to be constant, we can apply third equation of kinematics.

❍ v² - u² = 2as

where,

v denotes final velocity

u denotes initial velocity

a denotes acceleration

s denotes distance

Conversion :

↗ 1kmph = 1000/3600 = 5/18mps

↗ 30kmph = 30×5/18 = 8.33mps

Calculation :

→ v² - u² = 2as

→ (0)² - (8.33)² = 2(-0.5)s

→ s = (8.33)²/2(0.5)

→ s = 69.38m

[Note : -ve sign of a shows retardation.]

Answer 2

${ \rm{ \huge{ \red { \underline{ \underline{ Question:-}}}}}}$




▪ A bike is moving at a speed of 30 km/hr . Brakes are applied in order to provide uniform retardation of 0.5 m/sec ^2 . Find the distance covered bstore it stops?




${ \rm{ \red{ \huge{ \underline{ \underline{Solution:-}}}}}}$




${ \dagger{ \bold{ \purple{ \: \: \: GIVEN- }}}}$




${ \star{ \sf{ \blue{ \: \:initial \: speed \: the \: bike \: = 30 \: km \: {hr}^{ - 1}}}}} \\ \\ { \blue{ \sf{ = 30 \times \frac{5}{18} m \: {sec}^{ - 1}}}} \\ \\ { \blue{ \sf{ = \frac{25}{3} m \: {sec}^{ - 1}}}}$




${ \star{ \blue{ \sf{ \: \: retardation = 0.5 \: m \: {sec}^{ - 2}}}}}$




since, the brakes finally stop the bike...




${ \star{ \blue{ \sf{ \: \: final \: velocity \: of \: the \: bike = 0}}}}$



${ \dagger{ \bold{ \purple{ \: \: TO \: FIND- }}}}$





▪ Distance covered by the bike before it stops???



here,



using the third equation of motion......




${ \boxed{ \boxed{ \sf{ \pink{ \: \: \: {v}^{2} = {u}^{2} + 2aS \: \: \: }}}}}$



where,



${ \purple{ \sf{v = final \: velocity}}} \\ \\ { \purple{ \sf{u = initial \: velocity}}} \\ \\ { \purple{ \sf{a = acceleration}}} \\ \\ { \purple{ \sf{S = distance \: covered}}}$




▪ putting the above given values in the formula...



since , it's given in the question that the bike retards , the value of retardation is used with negative sign in the formula for acceleration..




${{ \sf{ \red{ {0}^{2} = {( \frac{25}{3} m \: {sec}^{ - 1} )}^{2} + 2( - 0.5 \: m \: {sec}^{ - 2} )S }}}}$




${ \implies{ \sf{ \red{0 = \frac{625}{ 3} {m}^{2} \: {sec}^{ - 2} -( 1m \: {sec}^{ - 2} )S}}}}$




${ \implies{ \red{ \sf{S \: \times \: 1 {m}^{2} \: {sec}^{ - 2} = \frac{625}{3} {m}^{2} \: {sec}^{ - 2} }}}}$




${ \implies{ \red{ \sf{S = \frac{ \frac{625}{9} {m}^{2} \: {sec}^{ - 2} }{1 m \: {sec}^{ - 2} } }}}}$




${ \implies{ \red{ \sf{S = \frac{625}{9} m}}}}$




${ \implies{ \boxed{ \boxed{ \green{ \sf{distance = 69.44 m}}}}}}$