Question

A car travelling at a velocity of 54 km / h , brakes to stop at the rate of - 0.25 m / s^2 . How many meters does it travel forward before coming to rest ?

111
113
112.5
225​

Answer 1

Given :

⟶ Initial velocity = 54kmph

⟶ Retardation = -0.25m/s²

⟶ Final velocity = zero

To Find :

➨ Distance covered by car.

SoluTion :

➳ Since acceleration has said to be constant throughout the motion, we can easily apply equation of kinematics to solve this question.

➳ Distance convered by body before it is brought to rest is called as stopping distance.

✴ Third equation of kinematics :

• v² - u² = 2as

⇒ 1kmph = 5/18mps

⇒ 54kmph = 54 × 5/18 = 15mps

$\implies\tt\:v^2-u^2=2as$

$\implies\tt\:(0)^2-(15)^2=2(-0.25)s$

$\implies\tt\:-225=-0.5s$

$\implies\tt\:s=\dfrac{225}{0.5}$

$\implies\underline{\boxed{\bf{s=450\:m}}}$

Answer 2

Given : –

• Initial velocity of the car (u) = 54 km/h = 15 m/s .
• Acceleration (a) = - 0.25 m/s² .
• Final velocity of the car (v) = 0 .

To Find : –

• Distance covered by the car before coming to rest (s) .

Solution : –

From 3rd equation of motion –

★ v² = u² + 2as

⪼ v² - u² = 2as

⪼ (0)² - (15)² = 2×(-0.25)s

⪼ -255 = -0.5s

⪼ s = -255 / -0.5

⪼ s = 450 m

Hence, the car travels 450m before coming to rest .