Question

A car start from rest and acquire a velocity of 72km/h in 20 second. Find 1.The acceleration 2. distance travelled by car assume motion of car is uniform?​

Answer 1

Velocity=72*1000/3600=20m/sec
a.
V=u+at
a=v/t=20/20= 1m/sec^2
b.
V^2-u^2=2as
S=v^2/2a= 20*20/2*1=200m

Answer 2

⇝ Given :-

• A car start from rest and acquire a velocity of 72km/h in 20 second.

⇝ To Find :-

• Acceleration of the Car.

• Distance Traveled By the Car.

⇝ Solution :-

❒ Converting Final Velocity in m/s :

We Have,

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

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

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

Here,

• Initial Velocity = u = 0 m/s

• Final Velocity = v = 20 m/s

• Time Taken = t = 20 s

1 》 Calculating Acceleration :-)

Let Acceleration of Bus 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 ;

$:\longmapsto20 = 0 + \text a \times 20$

$:\longmapsto20\text a = 20$

$:\longmapsto\text a = \cancel \frac{20}{20}$

$\purple{ \Large :\longmapsto  \underline {\boxed{{\bf a = 1\:m/s² } }}}$

Hence,

$\large\underline{\pink{\underline{\frak{\pmb{\text Acceleration = 1 \: m/s {}^{2} }}}}}$

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

$:\longmapsto\text{s} = 0 + \frac{1}{2} \times 1 \times {20}^{2}$

$:\longmapsto\text{s} = \cancel\frac{400}{2}$

$\purple{ \Large :\longmapsto  \underline {\boxed{{\bf s = 200 \: m} }}}$

Hence,

$\large\underline{\pink{\underline{\frak{\pmb{\text Distance \:\text Traveled = 200\: m }}}}}$