Question

The brakes of a car can produce a constant deceleration of 2m/s². Ashtey application of brakes the vehicle comes to rest in 9s. Fine the stopping distance.​

Answer 1

To Find:-

• Find the stoping distance.

Given:-

• The vehicle Congress to rest in 9 sec with an deceleration of 2m/s².

Solution:-

Here,

• u = initial velocity
• v = final velocity
• a = acceleration
• s = distance
• t = time

We have to use first equation:-

$\tt\implies \: v = u + at$

$\tt\implies \: 0 = u + ( -2 )( 9 )$

$\tt\implies \: 0 = u - 18$

$\tt\implies \: u = 18m/s$

Now,

$\tt\implies \: s = ut + \dfrac{1}{2} \times {at}^{2}$

$\tt\implies \: s = 18(9) + \dfrac{1}{2} \times -2 \times {9}^{2}$

$\tt\implies \: s = 162 - 81$

$\tt\implies \: s = 81m$

Answer 2

To Find:-

Find the stoping distance.

Given:-

The vehicle Congress to rest in 9 sec with an deceleration of 2m/s².

Solution:-

Here,

u = initial velocity

v = final velocity

a = acceleration

s = distance

t = time

We have to use first equation:-

$\tt\implies \: v = u + at$

$\tt\implies \: 0 = u + ( -2 )( 9 )$

$\tt\implies \: 0 = u - 18$

$\tt\implies \: u = 18m/s$

Now,

$\tt\implies \: s = ut + \dfrac{1}{2} \times {at}^{2}$

$\tt\implies \: s = 18(9) + \dfrac{1}{2} \times -2 \times {9}^{2}$

$\tt\implies \: s = 162 - 81$

$\tt\implies \: s = 81m$