Question

A vehicle is running at a speed of 45km/h.If it is stopped in 3 seconds by applying brakes,calculate the retardation of the vehicle and the distance travelled before it stopped.

Answer 1

Answer

• Retardation of the vehicle = 4.167 m/s²
• Distance travelled is 18.748 m

Given

• A vehicle is running at a speed of 45 km/h.If it is stopped in 3 seconds by applying brakes

To Find

• Retardation of the vehicle
• Distance travelled before it stopped

Concept Used

• Acceleration and Retardation are both same but opposite in direction .

Equations of motion :

• v = u + at

• s = ut + ¹/₂ at²
• v² - u² = 2as

Solution

Initial velocity , u = 45 km/h = 45 × ⁵/₁₈ m/s = 12.5 m/s

Time , t = 3 s

Final velocity , v = 0 m/s [ ∵ vehicle finally stops ]

Acceleration = - ? m/s² [ ∵ Retardation ]

Apply 1 st equation of motion .

⇒ v = u + at

⇒ 0 = 12.5 + a(3)

⇒ 3a = - 12.5

⇒ a = - 4.167 m/s²

So ,

Retardation = - a = 4.167 m/s²

______________________

Distance , s = ? m

Apply 3 rd equation of motion .

⇒ v² - u² = 2as

⇒ 0² - (12.5)² = 2( - 4.167 )s

⇒ - 156.25 = - 8.334 s

⇒ s = 18.748 m

Distance travelled is 18.748 m