Question

A car accelerates uniformly from 36 km/h to 72/h in 6 sec calculate the acceleration and the distance moved by the car in that time.​

Answer 1

⇝ Given :-

A car accelerates from 36 km/h to 72 km/h in 6 second.

⇝ To Find :-

Acceleration of the Car.

Distance Traveled By the Car.

⇝ Solution :-

❒ Converting Velocities in m/s :

We Have,

${Initial Velocity = 36 km/h}$

$\bigg(36 \times \frac{5}{18} \bigg) \text{m/s}$

$\red{:\longmapsto \text{Initial Velocity = 10 \: m/s}}$

Also,

${Final Velocity = 72 km/h}$

$\bigg(72 \times \frac{5}{18} \bigg) \text{m/s}$

$\red{:\longmapsto \text{Final Velocity = 20 \: m/s}}$

Here,

Initial Velocity = u = 10 m/s

Final Velocity = v = 20 m/s

Time Taken = t = 6 s

1 》 Calculating Acceleration :-)

Let Acceleration of Car be = a m/s².

❒ We Have 1st Equation of Motion as :

$\large\bf\red{v = u + at}$

Converting Velocities in m/s :

We Have,

$\text{Initial Velocity = 36 km/h}$

Applying 1st Equation of Motion ;

$\red {: \longmapsto 20 = 10 + \text a \times 6 }$

$\red{:\longmapsto6\text a = 20 - 10 }$

$\red{:\longmapsto6\text a = 10 }$

$\red{:\longmapsto\text a = \frac{10}{6}}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf a = 1.67 \: m/s {}^{2} } }}}$

Hence,

$\large\underline{\pink{\underline{\frak{\pmb{\text Acceleration = 1.67 \: m/s {}^{2} }}}}}$

2 》Calculating Distance Traveled 

❒ We Have 2nd Equation of Motion as :

$\large \bf \red \bigstar \purple{{ \underline{s=ut+\dfrac{1}{2}at^2}}}$

$\huge \red \bigstar$

Applying 2nd Equation of Motion ;

$\pink { \longmapsto \text{s} = 10 \times 6 + \frac{1}{2} \times \frac{10}{6} \times {6}^{2} }$

$\pink{:\longmapsto\text{s} = 60 + 30 }$

$\green{ \huge :\longmapsto  \underline {\boxed{{\bf s= 90 \: m} }}}$

Hence,

$\large\underline{\pink{\underline{\frak{\pmb{\text Distance \:\text Traveled = 90\: m }}}}}$

Answer 2

$\huge{\bf{\underline{\underline{ \colorbox{lavender}{\color{cyan}{αиѕωєя}}}}}}$

$⇝ Given :-$

A car accelerates from 36 km/h to 72 km/h in 6 second.

$⇝ To Find :-$

Acceleration of the Car.

Distance Traveled By the Car.

$⇝ Solution :-$

$❒ Converting Velocities in m/s :$

We Have,

${Initial \: Velocity = 36 km/h}$

$\bigg(36 \times \frac{5}{18} \bigg) \text{m/s}$

$\red{:\longmapsto \text{Initial Velocity = 10 \: m/s}}$

Also,

${Final Velocity = 72 km/h}$

$\bigg(72 \times \frac{5}{18} \bigg) \text{m/s}$

$\red{:\longmapsto \text{Final Velocity = 20 \: m/s}}$

Here,

Initial Velocity = u = 10 m/s

Final Velocity = v = 20 m/s

Time Taken = t = 6 s

•  Calculating Acceleration :-)

$Let \: Acceleration \: of \: Car \: be = a m/s².$

$❒ We \: Have \: 1st \: Equation \: of \: Motion \: as :$

$\large\bf\red{v = u + at}$

• Converting Velocities in m/s :

We Have,

$\text{Initial Velocity = 36 km/h}$

Applying 1st Equation of Motion ;

$\red {: \longmapsto 20 = 10 + \text a \times 6}$

$\red{:\longmapsto6\text a = 20 - 10 }$

$\red{:\longmapsto6\text a = 10 }$

$\red{:\longmapsto\text a = \frac{10}{6}}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf a = 1.67 \: m/s {}^{2} } }}}$

Hence,

$\large\underline{\pink{\underline{\frak{\pmb{\text Acceleration = 1.67 \: m/s {}^{2} }}}}}$

• 2 》Calculating Distance Traveled 

❒ We Have 2nd Equation of Motion as :

$\large \bf \red \bigstar \purple{{ \underline{s=ut+\dfrac{1}{2}at^2}}}$

• Applying 2nd Equation of Motion ;

$\pink { \longmapsto \text{s} = 10 \times 6 + \frac{1}{2} \times \frac{10}{6} \times {6}^{2} }$

$\pink{:\longmapsto\text{s} = 60 + 30 }$

$\green{ \huge :\longmapsto  \underline {\boxed{{\bf s= 90 \: m} }}}$

Hence,

$\large\underline{\pink{\underline{\frak{\pmb{\text Distance \:\text Traveled = 90\: m }}}}}$