Question

A scooter is running at a speed of 54km/h. Its velocity increases to 63km/h in next

one second. Calculate the acceleration acting on the scooter and the distance covered

by it in one second.​

Answer 1

Given :

• Initial velocity = 54 km/h
• Final Velocity = 63 km/h
• Time = 1 s

To Find :

• Acceleration produced
• Distance covered

Solution :

⠀⠀⠀⠀⠀⠀Acceleration of the Scooter :

First let us convert the unit's in m/s from km/h

• u = 54 km/h

→ (54 × 5/18) m/s

→ (3 × 5) m/s

→ 15 m/s

Hence, the initial velocity in m/s is 15 m/s.

• v = 63 km/h

→ (315/18) m/s

→ 17.5 m/s

Hence, the final Velocity in m/s is 17.5 m/s

To find the acceleration produced by the scooter , we can use the First Equation of Motion .

We know the , First Equation of Motion .i.e,

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

Where :-

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

Using the above equation and substituting the values in it , we get :

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

$:\implies \bf{17.5 = 15 + a \times 1}$

$:\implies \bf{17.5 = 15 + a}$

$:\implies \bf{17.5 - 15 = a}$

$:\implies \bf{2.5 = a}$

$\underline{\therefore \bf{Acceleration\:(a) = 2.5\:ms^{-2}}}$

Hence, the acceleration produced is 2.5 m/s².

⠀⠀⠀⠀⠀⠀⠀Distance covered in 1 s :

Method..i :-

To find the distance covered by the scooter , we can use the second Equation of Motion .

We know the , second equation of Motion .i.e,

$\underline{\boxed{:\implies \bf{S = ut + \dfrac{1}{2}at^{2}}}}$

Where :-

• S = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken

Using the second Equation of the motion and substituting the values in it, we get :

$:\implies \bf{S = ut + \dfrac{1}{2}at^{2}}$

$:\implies \bf{S = 15 \times 1 + \dfrac{1}{2} \times 2.5 \times 1^{2}}$

$:\implies \bf{S = 15 + \dfrac{1}{2} \times 2.5}$

$:\implies \bf{S = 15 + 1.25}$

$:\implies \bf{S = 16.25}$

$\underline{\therefore \bf{Distance\:(S) = 16.25\:m}}$

Hence, the distance covered in 1 s is 16.25 m.

Method..ii :-

To find the distance covered by the scooter , we can use the third Equation of Motion .

We know the , third equation of Motion .i.e,

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

Where :-

• S = Distance covered
• v = Final Velocity
• u = Initial velocity
• a = Acceleration

using the formula and substituting the values in it, we get :-

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

$:\implies \bf{17.5^{2} = 15^{2} + 2 \times 2.5 \times S}$

$:\implies \bf{17.5^{2} = 15^{2} + 5 \times S}$

$:\implies \bf{17.5^{2} - 15^{2} = 5 \times S}$

$:\implies \bf{81.25 = 5 \times S}$

$:\implies \bf{\dfrac{81.25}{5} = S}$

$:\implies \bf{16.25 = S}$

$\underline{\therefore \bf{Distance\:(S) = 16.25\:m}}$

Hence, the distance covered in 1 s is 16.25 m.