Question

a motor car is moving with a velocity of 108km/h and it takes 4s to stop after brakes applied .calculate the force exerted by the brakes on the motorcycle if its mass along with passengers is 1000kg​

Answer 1

$\large \dag$ Question :-

A motor car is moving with a velocity of 108km/h and it takes 4s to stop after brakes applied . Calculate the force exerted by the brakes on the motorcycle if its mass along with passengers is 1000kg.

$\large \dag$ Answer :-

$\red\dashrightarrow\underline{\underline{\sf  \green{Force   \: Exerted \:  is \:  -7500\: N }} }$

$\large \dag$ Step by step Explanation :-

❒ Converting Initial Velocity in m/s :-

We Have,

$\text{Initial Velocity = 108 \: km/h}$

$\rm = \bigg( 108 \times \frac{5}{18} \bigg) \: m/s$

$\red{:\longmapsto \rm Initial \: Velocity = 30 \: m/s }$

Now Here we have :

• Initial Velocity = u = 30 m/s

• Final Velocity = v = 0 m/s

• Time = t = 4 s

• Mass of motor car = m = 1000 kg

• Let Acceleration be = a m/s²

❒ We Have 1st Equation of Motion as :

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{v = u + at}}}$

⏩ Applying 1st Equation of Motion ;

$:\longmapsto \rm 0 = 30 + a \times 4$

$:\longmapsto \rm 0 = 30 + 4a$

$:\longmapsto \rm 4a= - 30$

$:\longmapsto \rm a = \frac{ - 30}{ \: \: 4}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf a = -7.5} }}}$

⇝ Acclerration is - 7.5 m/s²

We Know that

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{  \blue{F = ma }}}}$

Putting Values we get Force Exerted by Brakes is ;

$:\longmapsto \rm F = 1000 \times (- 7.5 )$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf F =-7500\: N} }}}$

Therefore,

$\large\underline{\pink{\underline{\frak{\pmb{\text Force \:  Exerted = -7500 \: Newton }}}}}$

✧ Negative Sign shows that force is exerted against the motion .

Answer 2

Given : A motor car is moving with a velocity of 108km/h and it takes 4s to stop after brakes applied .calculate the force exerted by the brakes on the motorcycle if its mass along with passengers is 1000kg

Question states that a car is moving with a uniform velocity (u) of 108 km/h after which takes 4s to stop. As the car stopped the final velocity (v) will be 0 m/sec. Total mass of the passengers is 1000 kg or 10 quintals

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Given Data :

Uniform Velocity (u) = 108 km/h ⟼ 108 × 5/18 ⟼ 30 m/sec

As the final velocity is in m/sec we will convert either of the two velocity. More applicable is to convert uniform velocity into m/sec. So, conversion is

◉ 1 km/h = 5/18 m/sec

Final Velocity (v) = 0 m/sec

Time taken (t) = 4 sec

Mass (m) = 1000 kg

Using Newton's Second Law of Motion

❍ F = ma

Here,

F represents force

m represents mass

a represents acceleration

$\star \quad \frak{F = ma } \\ \\ [ \sf Note : \frak{Acceleration \: can \: be \: broken \: down \: into \: \green{( \frac{v - u}{t})}}] \\ \\ \dashrightarrow \frak{F = 1000 \times \frac{0 - 30}{4} } \\ \\ \dashrightarrow \frak{F = 1000 \times( - \frac{30}{4}) } \\ \\ \dashrightarrow \frak{F = 1000 \times ( - 7.5) } \\ \\ \star \quad \underline{ \boxed{ \pink{ \frak{F = - 7500 N}}}}$

• Henceforth, the Force exerted is -7500 N