Question

3. A car is travelling at a speed of 90 km/h. Brakes are applied so as to produce a
uniform acceleration of - 0.5 m/s2. Find how far the car will go before it is brought to
rest?​

Answer 1

Given :-

A car is travelling at a speed of 90 km/hr . Brakes are applied so as to produce an uniform acceleration of - 0.5 m/s² .

Required to find :-

• How far the car will go before it is brought to rest ?

Equations used :-

• v = u + at

• s = ut + ½ at²

Solution :-

Given data :-

A car is travelling at a speed of 90 km/hr . Brakes are applied so as to produce an uniform acceleration of - 0.5 m/s² .

we need to find the distance covered by the car before it is brought to rest .

So,

From the given data we can conclude that ;

• Initial velocity of the car = 90 km/hr

• Acceleration = - 0.5 m/s²

• Final velocity of the car = 0 km/hr

Before solving this question we need to convert the velocities from km/hr to m/s . Since, acceleration is given in m/s²

So,

1 km/hr = 5/18 m/s

90 km/hr = ? m/s

=> 90 x 5/18

=> 25 m/s

similarly,

0 km/hr = 0 m/s

Hence,

Initial velocity of the car = 25 m/s

Final velocity of the car = 0 m/s

Using the equation of motion ;

i.e. v = u + at

0 = 25 + ( - 0.5 x t )

0 = 25 + ( - 0.5t )

- 25 = - 0.5t

Taking - ( minus ) common on both sides ;

- ( 25 ) = - ( 0.5t )

- ( minus ) get's cancelled

25 = 0.5t

0.5t = 25

t = 25/0.5

t = 250/5

t = 50 seconds

Hence,

Time taken = 50 seconds

Now,

Using the 2nd equation of motion ;

i.e. s = ut + ½ at²

s = 25 x 50 + ½ x - 0.5 x 50 x 50

s = 1250 + ½ x - 0.5 x 50 x 50

s = 1250 + - 0.5 x 25 x 50

s = 1250 + ( - 0.5 x 1250 )

s = 1250 + ( - 625 )

s = 1250 - 625

s = 625 meters

Hence,

• Displacement ( s ) = 625 meters

Therefore ;

The distance covered by the car before it is brought to rest is 625 meters.

Answer 2

Answer

• Car will go 625 meters before it is brought into rest

Given

• A car is travelling at a speed of 90 km/h. Brakes are applied so as to produce a  uniform acceleration of - 0.5 m/s² .

To Find

• How far the car will go before it is brought to  rest

Solution

Initial velocity , u = 90 km/h = 90 × ⁵/₁₈ m/s= 25 m/s

⇒ Initial velocity , u = 25 m/s

Acceleration , a = - 0.5 m/s²  

[ Here -ve sign denotes deceleration or retardation ]

Distance , s = ? m

Final velocity , v = 0 m/s

[ ∵ Finally car brought into rest ]

Here , v , u , a are given . We need to find s .

So , Use 3rd equation of motion .

⇒ v² - u² = 2as

⇒ 0² - 25² = 2( - 0.5 ) s

⇒ - 625 = - s

⇒ s = 625 m

More Info

Equations of motion :

• v = u + at
• s = ut + ¹/₂ at²
• v² - u² = 2as
• Sₙ = u + ᵃ/₂ [ 2n - 1 ]