Question

(a) Derive a mathematical expression for position-time relation i.e. S = ut + ½ at2. 5 (b) A car acquires a velocity of 72 Km/h in 10 seconds starting from rest. Find (i) Acceleration (ii) distance travelled in this time ​

Answer 1

Question :

To derive the mathematical expression of :

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

Solution :

To derive the above equation , we need a idea of Average Velocity.

We know that , Average Velocity is :

$\bf{Average\:Velocity = \dfrac{v + u}{2}}$ Eq.(i)

Where :-

• v = Final Velocity
• u = Initial Velocity

[Note : Here the acceleration is constant]

On the other case , we know that the Average Velocity is the ratio of the total distance and the total time.i.e,

$\bf{Average\:Velocity = \dfrac{s}{t}}$.(ii)

Where :-

• s = Total Distance
• t = Total time Taken

Now , on comparing the Equation (i) and Equation (ii) , we get :

$:\implies \bf{\dfrac{v + u}{2} = \dfrac{s}{t}}$

Now , using the first Equation of Motion .i.e,

⠀⠀⠀⠀⠀⠀⠀⠀⠀v = u + at.

And Substituting it in the equation , we get :

$:\implies \bf{\dfrac{(u + at) + u}{2} = \dfrac{s}{t}}$ [$\because$ v = u + at]

$:\implies \bf{\dfrac{2u + at}{2} = \dfrac{s}{t}}$

$:\implies \bf{\bigg(\dfrac{2u + at}{2}\bigg)t = s}$

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

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

$:\implies \bf{\dfrac{\not{2}ut}{\not{2}} + \dfrac{at^{2}}{2} = s}$

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

$\underline{\boxed{\therefore \bf{s = ut + \dfrac{at^{2}}{2}}}}$

Hence Derived !!

Question :

A car acquires a velocity of 72 Km/h in 10 seconds starting from rest. Find

⠀⠀⠀⠀⠀⠀(i) Acceleration

⠀⠀⠀⠀⠀⠀(ii) distance travelled in this time

Given :

• Final Velocity = 72 km/h
• Time = 10 s
• Initial velocity = 0

To find :

• Acceleration of the car.

• Distance traveled.

Solution :

⠀⠀⠀⠀⠀⠀To Find the Acceleration :

First let us convert the final Velocity in m/s .

To convert the final Velocity in m/s from km/h , multiply it by 5/18.

• v = 72 km/h

→ (72 × 5/18) m/s

→ (4 × 5) m/s

→ 20 m/s

Hence, the final Velocity in m/s is 20 m/s.

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

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

$:\implies \bf{20 = 0 + a \times 5}$

$:\implies \bf{20 = a \times 5}$

$:\implies \bf{\dfrac{20}{5} = a}$

$:\implies \bf{4 = a}$

$\therefore \bf{a = 4\:ms^{-2}}$

Hence, the acceleration produced is 4 m/s².

⠀⠀⠀⠀⠀⠀To find the Distance traveled :

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

$:\implies \bf{S = 0 \times 5 + \dfrac{1}{2} \times 4 \times 5^{2}}\:\:(u = 0)$

$:\implies \bf{S = \dfrac{1}{\not{2}} \times \not{4}\times 25}$

$:\implies \bf{S = 2 \times 25}$

$:\implies \bf{S = 50}$

$\therefore \bf{S = 50\:m}$

Hence, the distance traveled is 50 m.