Question

A car moving with a speed of 15 m/s can be stopped by applying
brakes, after travelling a distance of 3m. The retardation of the car is
1) 37.5m/s 2) 47.5m/s. 3) 57.5m/s
4) 67.5m/s​

Answer 1

Given :-

A car moving with a speed of 15 m/s can be stopped by applying brakes after travelling a distance of 3 meters .

Required to find :-

• The retardation of the car ?

Equation used :-

1. v² - u² = 2as

Solution :-

Given information :-

A car moving with a speed of 15 m/s can be stopped by applying brakes after travelling a distance of 3 meters .

we need to find the retardation of the car

So,

From the given information we can conclude that ;

• Initial velocity of the car ( u ) = 15 m/s

• Final velocity of the car ( v ) = 0 m/s

• Distance travelled ( s ) = 3 meters

Using the equation of motion ;

✎ v² - u² = 2as

✎ ( 0 )² - ( 15 )² = 2 x a x 3

✎ 0 - 225 = 6a

✎ - 225 = 6a

✎ 6a = - 225

✎ a = - 225/6

✎ a = - 37.5 m/s²

Hence,

Acceleration = - 37.5 m/s²

Since negative acceleration is termed as " retardation "

Therefore,

Retardation of the car ( a ) = - 37.5 m/s²

Additional information :-

The other equations of motion are ;

v = u + at

s = ut + ½ at²

These equations are applicable when the acceleration is constant .

The letter s represents displacement

The most important to be remembered is we can take displacement in the terms of distance but we can't take distance in terms of displacement .

Speed is a scalar quantity .

Velocity is a vector quantity .

Speeds needs only magnitude but velocity needs both magnitude and direction .

Answer 2

Given:-

• Initial velocity (u) of car = 15m/s

• Final velocity (v) of car = 0m/s

• Distance (s) traveled by car = 3 m.

To Find :-

• Retardation (a) of the car

Solution :-

We have given the initial velocity of car, final velocity of car and distance covered by car .we have to find the retardation of car.

So, By using 3rd equation of motion

➦ v² = u² + 2as

Putting all the values which is given above

➲ 0² = 15² + 2 ×a ×3

➲ 0 = 225 + 6×a

➲ -225 = 6× a

➲ a = -225/6

➲ a = -37.5 m/s²

here negative sign show retardation

∴ The retardation of car is 37.5 m/s²

Hence, option (a) 37.5 m/s² is the correct answer.

Additional Information

☛ The rate of change of velocity at per unit time is called acceleration.

☛ Negative acceleration is called Retardation.

☛ It is vector quantity.

☛ SI unit is m/s²