Question

A car at arts from rest and aquire a velocity of 54km/hr in 2 sec find distance travelled by the car assumed that the motion of the car is uniform

Answer 1

Given :

• Final Velocity of the car = 54 km h¯¹

• Time Taken = 2s

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

To Find :

⠀⠀⠀⠀⠀The distance covered by the car.

Solution :

To find the distance covered by the car first we have to find the acceleration produced by the car.

But before finding the acceleration , let us make all the units same in all the conditions.

Given that the final Velocity of the car is 54 km h¯¹ and the time is 2 s , so either we have to convert the velocity in m s¯¹ or the time in h .

So , let us convert the final Velocity in m s¯¹ from km h¯¹.

To convert the unit from km h¯¹ to m s¯¹ we have to multiply it by 5/18 , so we get :

$\boxed{\begin{minipage}{4 cm} }bf{v = 54 km h^{-1}} \\ \\ :\implies \bf{v = \bigg(54 \times \dfrac{5}{18}\bigg) m s^{-1}} \\ \\ :\implies \bf{v = (3 \times 5) m s^{-1}} \\ \\ :\implies \bf{v = 15 m s^{-1}} \end{minipage}}$

Hence, the final Velocity in m s¯¹ is 15 m s¯¹.

Now :

⠀⠀⠀⠀⠀⠀⠀⠀⠀To find the acceleration :

Using the First Equation of Motion and substituting the values in it, we get :-

$:\implies \underline{\bf{v = u + at}}$

Where :-

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = time taken

$:\implies \bf{15 = 0 + a \times 3}\:\:\:(\because u = 0)$

[Note : The Initial Velocity will be zero as the car started from rest]

$:\implies \bf{15 = 3a}$

$:\implies \bf{\dfrac{15}{3} = a}$

$:\implies \bf{5 = a}$

$\therefore \bf{a = 5 ms^{-2}}$

Hence, the acceleration produced by the car is 5 m s¯².

⠀⠀⠀⠀⠀⠀⠀To find the distance traveled :

Using the Third Equation of Motion and substituting the values in it, we get :

$:\implies \underline{\bf{v^{2} = u^{2} + 2as}}$

Where :-

• v = Final Velocity
• u = initial velocity
• a = Acceleration
• s = Distance covered

$:\implies \bf{15^{2} = 0^{2} + 2 \times 5 \times s}\:\:\:(\because u = 0)$

$:\implies \bf{225 = 10 \times s}$

$:\implies \bf{\dfrac{225}{10} = s}$

$:\implies \bf{22.5 = s}$

$\therefore \bf{Distance = 22.5 m}$

Hence, the distance covered by the car is 22.5 m.