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,

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

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

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

Also,

$\text{Final Velocity = 72 km/h}$

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

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

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 :

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

⏩ Applying 1st Equation of Motion ;

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

$:\longmapsto6\text a = 20 - 10$

$:\longmapsto6\text a = 10$

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

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

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 :

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

⏩ Applying 2nd Equation of Motion ;

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

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

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf s= 90 \: m} }}}$

Hence,

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

Answer 2

Answer:

Given :-

• A car accelerates uniformly from 36 km/h to 72 km/h in 6 seconds.

To Find :-

• What is the acceleration.
• What is the distance moved by the car in that time.

Formula Used :-

$\clubsuit$ First Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

$\clubsuit$ Second Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2}at^2}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken
• s = Distance Covered

Solution :-

First, we have to convert the initial and final velocity km/h into m/s :

❒ In case of Initial Velocity (u) :

$\implies \sf Initial\: Velocity =\: 36\: km/h$

$\implies \sf Initial\: Velocity =\: 36 \times \dfrac{5}{18}\: m/s\: \bigg\lgroup \sf\bold{\pink{1\: km/h =\: \dfrac{5}{18}\: m/s}}\bigg\rgroup$

$\implies \sf Initial\: Velocity =\: \dfrac{180}{18}\: m/s$

$\implies \sf\bold{\purple{Initial\: Velocity =\: 10\: m/s}}$

❒ In case of Final Velocity (v) :

$\implies \sf Final\: Velocity =\: 72\: km/h$

$\implies \sf Final\: Velocity =\: 72 \times \dfrac{5}{18}\: m/s\: \: \bigg\lgroup \sf\bold{\pink{1\: km/h =\: \dfrac{5}{18}\: m/s}}\bigg\rgroup$

$\implies \sf Final\: Velocity =\: \dfrac{360}{18}\: m/s$

$\implies \sf\bold{\purple{Final\: Velocity =\: 20\: m/s}}$

Now, we have to find the acceleration of the car :

Given :

• Final Velocity (v) = 20 m/s
• Initial Velocity (u) = 10 m/s
• Time Taken (t) = 6 seconds

According to the question by using the formula we get,

$\longrightarrow \sf 20 =\: 10 + a(6)$

$\longrightarrow \sf 20 =\: 10 + 6a$

$\longrightarrow \sf 20 - 10 =\: 6a$

$\longrightarrow \sf 10 =\: 6a$

$\longrightarrow \sf \dfrac{\cancel{10}}{\cancel{6}} =\: a$

$\longrightarrow \sf 1.67 =\: a$

$\longrightarrow \sf\bold{\red{a =\: 1.67\: m/s^2}}$

${\small{\bold{\underline{\therefore\: The\: acceleration\: of\: the\: car\: is\: 1.67\: m/s^2\: .}}}}$

Now, we have to find the distance moved by the car in that time :

Given :

• Initial Velocity (u) = 10 m/s
• Time Taken (t) = 6 seconds
• Acceleration (a) = 1.67 m/s²

According to the question by using the formula we get,

$\longrightarrow \sf s =\: (10)(6) + \dfrac{1}{2} \times (1.67)(6)^2$

$\longrightarrow \sf s =\: 10 \times 6 + \dfrac{1}{2} \times 1.67 \times (6 \times 6)$

$\longrightarrow \sf s =\: 60 + \dfrac{1}{2} \times 1.67 \times 36$

$\longrightarrow \sf s =\: 60 + \dfrac{1}{2} \times 60.12$

$\longrightarrow \sf s =\: 60 + 30.06$

$\longrightarrow \sf s =\: 90.06$

$\longrightarrow \sf s \approx 90$

$\longrightarrow \sf\bold{\red{s =\: 90\: m}}$

${\small{\bold{\underline{\therefore\: The\: distance\: moved\: by\: the\: car\: is\: 90\: m\: .}}}}$