Question

a car is moving with a speed of 50km/h if driver applies brakes and car stops within 10 metres find the deceleration

Answer 1

Answer

• Deceleration = 9.465 m/s²

Given

• A car is moving with a speed of 50 km/h . Driver applies brakes and car stops within 10 metres

To Find

• Deceleration

Concept Used

• Acceleration is nothing but deceleration or retardation but opposite in direction .
• SI unit acceleration is m/s²

Equation of motion ( 3 )

$\bigstar \bf \;\; v^2-u^2=2as$

Solution

Initial velocity , u = 50 km/h = 50 × ⁵/₁₈ = 13.89 m/s²

Final velocity , v = 0 m/s [ ∵ Finally stops after applying brakes ]

Distance , s = 10 m

Acceleration , a = ? m/s²

Apply 3 rd equation of motion .

⇒ v² - u² = 2as

⇒ 0² - (13.89)² = 2a(10)

⇒ - 192.9 = 20a

⇒ 20a = - 192.9

⇒ a = - 9.645 m/s²

Note : - ve sign of acceleration denotes deceleration / retardation .

So , Deceleration = 9.645 m/s²

Answer 2

• Given -

a car is moving with a speed of 50km/h & driver applies brakes and car stops within 10 m

• To Find -

Deceleration (-) represents Deceleration or rate of decrease of acceleration in car

• Solution -

Here, Formula used is 3rd Equation Of Motion. As it's directly related to the term Acceleration.

☆ ${v}^{2} - {u}^{2} = 2as$

Here,

• V = Final Velocity
• U = Initial / starting Velocity
• A = Acceleration
• S = Distance

Changes according to given Conditions,

• V = 0 .. ( Brakes are applied & car stops. So, No Acceleration)
• U = 50 km/h .. (Conversion Of Speed) .. (50×5/18 = 13.89 m/s^2)
• A = We have to Find out
• S = 10m

Now, Let's apply those changes in our Equation,

$⟹ {v}^{2} - {u}^{2} = 2as$

$⟹ {0}^{2} - ( {13.89}^{2} ) = 2a(10)$

$⟹0 - 192.9 = 20a$

$⟹ - 192.9 = 20a$

$⟹20a = - 192.9$

$⟹a = \frac{\cancel{-1</strong><strong>9</strong><strong>2</strong><strong>.</strong><strong>9</strong><strong>}}{\cancel{20}}$

$⟹a = </strong><strong>-</strong><strong>9.645 {m</strong><strong>/</strong><strong>s}^{2}$

⛬ Deceleration is -9.645m/s^2