Question

Abus is moving with a velocity of 72 Kmh. On the application of the
breakes its stops after covering a distance of 500 m. Calculate the
deceleration produced by the breakes.​

Answer 1

Answer:

5.184ms^-2

Explanation:

u^2+2as=v^2

72^2+2a*500m=0

5184+2a*500m=0

2a= -5184/500

a=-5184/1000

a=-5.184ms^-2

negative acceleration is deceleration.

therefore deceleration is 5.184ms^-2

Answer 2

Given :

• Final velocity, v = 0 m/s
• Initial velocity, u = 72 km/h
• Distance covered, s = 500 m

To find :

• Deceleration produced by the brakes

According to the question,

★ Note :

After changing the Initial velocity into m/s. The initial velocity will be 20 m/s.

➞ v² = u² + 2as

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• s = Distance

➞ Substituting the values,

➞ (0)² = (20)² + 2 × a × 500

➞ 0 = 400 + 1000a

➞ 0 - 400 = 1000a

➞ - 400 = 1000a

➞ - 400 ÷ 1000 = a

➞ - 0.4 = a

• So, the deceleration produced by the brakes is 0.4 m/s².

____________________

★ Additional information :

• Negative acceleration is also known as retardation.

• Its SI unit is m/s².

• Acceleration : Rate of change of velocity per unit time is known as acceleration.

★ Newtons equation of motion :

First equation :

• v = u + at

Second equation :

• s = ut + ½ at²

Third equation :

• v² = u² + 2as