A motor car moving with a uniform speed of $20m{s}^{-1}$ comes to rest on the application of brakes after travelling a distance of 10 m. Its acceleration is

Step 1: Given that:

The initial speed of the car($u$) = $20m{s}^{-1}$

As the car comes to rest after brakes,

The final velocity of the car($v$) = 0

Distance travelled by the car(s) = $10m$

Step 2: Calculation of the acceleration of the car:

According to the third equation of motion;

$v^2=u^2+2as$

Where u = Initial velocity

v= Final velocity

s= Distance travelled by the body

a= Acceleration

Putting all the values, we get

$0={\left(20m{s}^{-1}\right)}^2+2a\times 10m$

$0=400+20a$

$\Rightarrow 400+20a=0$

$\Rightarrow 20a=-400$

$\Rightarrow a=-20m{s}^{-2}$

(Negative sign indicates that the acceleration in the car is negative that is the car has retardation.)

Thus,

The acceleration in the car is $-20m{s}^{-2}$