Question

A car starting from rest attains a velocity of 20 ms- in 5 seconds. Find
the distance covered by the car.​

Answer 1

Given :

• Initial velocity, u = 0 m/s (Starting from rest)
• Final velocity, v = 20 m/s
• Time taken, t = 5 seconds

To find :

• Distance covered by the car, s

According to the question,

➞ v = u + at

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• t = Time taken

➞ Substituting the values,

➞ 20 = 0 + a × 5

➞ 20 - 0 = 5a

➞ 20 = 5a

➞ 20 ÷ 5 = a

➞ 4 = a

• So, the acceleration is 4 m/s².

Now,

➞ s = ut + ½ at²

Where,

• s = Distance
• u = Initial velocity
• t = Time taken
• a = Acceleration

➞ s = 0 × 5 + ½ × 4 × 5 × 5

➞ s = 0 + 2 × 25

➞ s = 0 + 50

➞ s = 50

• So, the distance covered by the car is 50 metres.

____________________

Answer 2

$\Large\bf{\color{cyan}GiVeN,}$

• A car starting from rest, i.e. initial velocity of car is 0 m/s.

• It attains a velocity of 20 m/s, i.e. final velocity of car is 20 m/s.

• Time period is 5 s.

$\bf\blue{We\:know \:that,}$

$\red\bigstar\:\:\bf\purple{v\:=\:u\:+\:at\:}$

$\bf\pink{Where,}$

• v = final velocity = 20 m/s

• u = initial velocity = 0 m/s

• t = Time = 5 s

• a = acceleration = ?

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

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

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

$:\implies\:\:\bf\orange{a\:=\:4\:m/s^2}$

$\bf\green{Again\:we\:know \:that,}$

$\pink\bigstar\:\:\bf\blue{S\:=\:ut\:+\:\dfrac{1}{2}at^2\:}$

$\bf\red{Where,}$

• S = Distance covered by car

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

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

$:\implies\:\:\bf{\color{peru}S\:=\:50\:m\:}$

$\Large\bf\therefore$ The distance covered by car 50 m.