Question

A car is moving with speed 72 km/h on a straight road.
Suddenly brakes are applied due to which car retards
at the rate of 10 m/s?, distance travelled by car before
coming to rest is​

Answer 1

Answer:

• Distance covered by Car before coming to rest = 20 metres

Explanation:

Given:

• Initial velocity of Car, u = 72 km/h
• Car is running on a straight road
• On applying brakes Car retards so, acceleration of Car, a = -10 m/s²
• Car comes to rest so, final velocity of Car, v = 0  

To find:

• Distance travelled by Car before coming to rest, s =?

Formula required:

• Third equation of motion

         2 a s = v² - u²

[ Where a is acceleration, s is distance covered, v is final velocity, u is initial velocity ]

Solution:

We are given Initial velocity of Car

→ u = 72 km/h

→ u = 72 × 5/18  m/s

→ u = 4 × 5  m/s

→ u = 20 m/s

Using third equation of motion

→ 2 a s = v² - u²

→ 2 (-10) ( s ) = (0)² - (20)²

→ -20 s = 0 - 400

→ -20 s = -400

→ 20 s = 400

→ s = 400/20

→ s = 20 m

Therefore,

• Distance covered by Car before coming to rest will be 20 metres.

Answer 2

Explanation:

Given : -

• A car is moving with speed 72 km/h on a straight road.

• Suddenly brakes are applied due to which car retards at the rate of 10 m/s?,

To Find : -

• distance travelled by car before coming to rest is

We are given Initial velocity of Car

10 × 2 = 20 m/s

u = 72 km/h

u = 72 × 20/72 m/s

u = 20 m/s

we have

2 a s = v² - u²

substitute all values

2 (-10) ( s ) = (0)² - (20)²

- 20s = - 400

s = 400/ 20

s = 20 m

Hence the answer is 20 m