Question

. A car is travelling along the road at 8 m/s . It accelerates by 1 m/s² for a distance 18m. How fast is it then travelling?

Answer 1

Given :

➳ Initial velocity = 8m/s

➳ Acceleration = 1m/s²

➳ Distance travelled = 18m

To Find :

➾ Final velocity of car.

SoluTion :

➨ Since, Acceleration of car is said to be constant throughout the motion, we can easily apply equation of kinematics to solve this question.

✴ Third equation of kinematics :

• v² - u² = 2as

where,

◕ v denotes final velocity

◕ u denotes initial velocity

◕ a denotes acceleration

◕ s denotes distance

⇒ v² - u² = 2as

⇒ v² - (8)² = 2(1)(18)

⇒ v² - 64 = 36

⇒ v² = 100

⇒ v = √100

⇒ v = 10 m/s

Answer 2

▪ $\underline\mathrm{Given-}$

• $\underline\mathrm{Initial \: Velocity \: = 8m/s}$
• $\underline\mathrm{Acceleration = 1m/s^2}$
• $\underline\mathrm{Distance = 18m}$

▪ $\underline\mathrm{To \: Find-}$

• $\underline\mathrm{Velocity(speed) \: of \: car}$

▪ $\underline\mathrm{Solution-}$

We know 3rd Equation Of Motion,

★ V^2 - U^2 = 2 as

Here,

• $\underline\mathrm{Initial \: Velocity = u}$

• $\underline\mathrm{Final \: Velocity = v}$

• $\underline\mathrm{Acceleration = a}$

• $\underline\mathrm{Distance = s}$

Placing Values,

⟶ v^2 - (8)^2 = 2 × (1) × (18)

⟶ v^2 - 64 = 18 × 2

⟶ v^2 - 64 = 36

⟶ v^2 = 36 + 64

⟶ v^2 = 100

⟶ v^2 = √100

⟶ v^2 = 10 m/s

⛬ Velocity of car is 10m/s