Question

A car of mass 1000kg travelling at 72km/h is uniformly bought to rest over a distance of 40m .find (a) the average acceleration (b) the average breaking force
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Answer 1

Given:-

→ Mass of the car = 1000kg

→ Initial velocity = 72km/h

→ Distance travelled = 40m

To find:-

→ Average acceleration

→ Average breaking force

Solution:-

Firstly, let's convert the initial velocity of the car from km/h to m/s.

=> 1km/h = 5/18m/s

=> 72km/h = 5/18×72

=> 20m/s

In this case :-

• Final velocity of the car will be zero as it finally comes to rest.

Let's calculate the acceleration of the car by using the 3rd equation of motion :-

=> v² - u² = 2as

=> 0 - (20)² = 2×a×40

=> -400 = 80a

=> a = -400/80

=> a = -5m/s²

Now, let's calculate the breaking force by using Newton's 2nd law of motion :-

=> F = ma

=> F = 1000(-5)

=> F = -5000 N

[ Here, -ve sign represents breaking force ]

Thus:-

• The average acceleration is -5m/s² .

• The average breaking force is 5000N .