Question

(e) A car running at 20ms^(-1) slows down to 5ms^(-1) by the application of brakes over a distance of 30m .Calculate the deceleration of the car.​

Answer 1

Given:-

→ Initial velocity of the car = 20 m s⁻¹

→ Final velocity of the car = 5 m s⁻¹

→ Distance travelled = 30 m

To find:-

→ Deceleration [-ve acceleration] of the car.

Solution:-

By using the 3rd equation of motion, we get :-

v² - u² = 2as

Where :-

• v is final velocity of the body.

• u is initial velocity of the body.

• a is acceleration of the body.

• s is the distance travelled.

Substituting values, we get :-

⇒(5)² - (20)² = 2(a)(30)

⇒25 - 400 = 60a

⇒ -375 = 60a

⇒ a = -375/60

⇒ a = -6.25 m s⁻²

∵ Acceleration is -6.25 m s⁻²

∴ Deceleration of the car is 6.25 m s⁻² .

Answer 2

Answer:

Given :-

• A car running at 20 m/s-¹ slows down to 5 m/s-¹ by the application of brakes over a distance of 30 m.

To Find :-

• What is the deceleration of the car.

Formula Used :-

➲ $\sf\boxed{\bold{\pink{a =\: \dfrac{{v}^{2} - {u}^{2}}{2s}}}}$

where,

• a = Deceleration
• v = Final Velocity
• u = Initial Velocity
• s = Distance

Solution :-

Given :

• Final Velocity = 5 m/s-¹
• Initial Velocity = 20 m/s-¹
• Distance = 30 m

According to the question by using the formula we get,

↦ $\sf a = \dfrac{{(5)}^{2} - {(20)}^{2}}{2(30)}$

↦ $\sf a = \dfrac{25 - 400}{60}$

↦ $\sf a = \dfrac{\cancel{- 375}}{\cancel{60}}$

➠ $\sf\bold{\red{a =\: - 6.25\: m/{s}^{2}}}$

$\therefore$ The deceleration of a car is 6.25 m/s².