Question

a bus is travelling at a speed of 72 km per hour the driver applies brakes a uniform acceleration of 0.2 metre per second square is produced find out the distance travelled by the bus before it comes to rest

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Answer 1

Answer:

here the key focus areas are the -ve sign for acceleration and the formula v²-u²= 2aS

hope it helps...

Answer 2

$\huge\purple{ \underline{ \boxed{\mathbb{\red{QUESTION : }}}}}$

A train is travelling at a speed of 72 Km/h . The driver applies brakes so that a uniform acceleration of -0.2 ms^-2 is produced .Find the distance travelled by train before it comes to rest .

$\huge\purple{ \underline{ \boxed{\mathbb{\red{ANSWER : }}}}}$

• Distance travelled by train = 1 Km

$\huge\purple{ \underline{ \boxed{\mathbb{\red{EXPLANATION : }}}}}$

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• Initial Velocity ( u ) = 72 $\sf Km {h}^{ - 1}$

• Acceleration ( a ) = -0.2 $\sf m {s}^{ - 2}$

• Final Velocity ( v ) = 0 $\sf m {s}^{ - 1}$ ( At rest )

$\green{ \large \underline{ \mathbb{\underline{TO \: FIND : }}}}$

• Distance travelled by train ( s ) .

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$\purple{ \boxed{ \sf {v}^{2} - {u}^{2} = 2as }}$

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$\red{\textsf{ \underline{\underline{Converting initial velocity into SI unit : }}}}$

To convert kilometre per hour into metre per hour , we multiply the value by 5/18 .

$\displaystyle \sf \longrightarrow u = \bigg( 72 \times \frac{5}{18} \bigg)m {s}^{ - 1} \: \\ \\ \displaystyle \sf \longrightarrow u = \bigg( 4 \times 5 \bigg)m {s}^{ - 1} \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow u = 20 \: m {s}^{ - 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\red{\textsf{ \underline{\underline{Distance travelled by train ( s ) : }}}}$

$\displaystyle \sf \longrightarrow {v}^{2} - {u}^{2} = 2as \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow {(0)}^{2} - {(20)}^{2} = 2 \times ( - 0.2) \times s \: \\ \\ \displaystyle \sf \longrightarrow - 400 = - 0.4s \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow - 0.4s = - 400 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow s = \frac{ \cancel{-} 400}{ \cancel{-} 0.4} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow s = 1000 \: m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow s = 1 \: Km \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\purple{ \boxed{ \begin{array}{l} \textsf{Distance travelled by train ( s ) = 1 Km}\end{array}}}$

$\green{ \large \underline{ \mathbb{\underline{KNOW\:MORE: }}}}$

• Acceleration

Acceleration is the rate at which velocity changes with time . Negative acceleration is called retardation .

• Initial Velocity

Initial velocity is the velocity of the object before the effect of acceleration .

• Final Velocity

Final velocity is the velocity of the object after the effect of acceleration .

• Distance

Distance is the length of actual path covered by a moving object in a given time interval .

$\Large{\purple{\boxed{\begin{array}{l} \textsf{Equations of motion : } \\ \\ \textsf{v = u + at} \\ \\ \displaystyle\textsf{s = ut + \sf\frac{1}{2}a {t}^{2} } \\ \\ \sf {v}^{2} - {u}^{2} = 2as \end{array}}}}$