Question

The brakes applied to a car produce an acceleration of 6 m s-2 in the opposite direction to the motion. If the car takes 3 s to stop after the application of brakes, calculate the distance it travels during this time. Also calculate the initial velocity of the car

Answer 1

EXPLANATION.

=> The break applied to a car produce an

acceleration = 6 m/s².

=> it take 3 seconds to stop after the application

of breaks.

=> Let the acceleration = - 6 m/s².

=> time taken = 3 seconds.

=> final velocity = 0

1) = calculate the initial velocity of car.

=> From Newton 1 st equation of kinematics.

=> v = u + at

=> 0 = u - 6 X 3

=> u = 18 m/s.

2) = calculate the distance it travel

during the time.

From the Newton 2nd equation of kinematics.

=> s = ut + ½ at²

=> s = 18 X 3 + ½ X (-6) X 3 X 3

=> s = 54 - 27

=> s = 27 m

Answer 2

Given :

• Acceleration = -6m/s²
• Time Taken = 3 seconds
• Final Velocity = 0m/s

To Find :

• Initial Velocity
• Distance Travelled.

Solution :

Firstly we will find Initial Velocity :

Using 1st Equation :

$\longmapsto\tt\boxed{v=u+at}$

Putting Values :

$\longmapsto\tt{0=u+(-6)(3)}$

$\longmapsto\tt{0=u-18}$

$\longmapsto\tt\bold{u=18m/s}$

So , The Initial Velocity of the car is 18m/s...

Now ,

For distance :

Using 2nd Equation :

$\longmapsto\tt\boxed{s=ut+\dfrac{1}{2}a{t}^{2}}$

Putting Values :

$\longmapsto\tt{s=18(3)+\dfrac{1}{2}(-6){(3)}^{2}}$

$\longmapsto\tt{s=54+(-3)\times{(9)}}$

$\longmapsto\tt{s=54-3-9}$

$\longmapsto\tt{s=54-27}$

$\longmapsto\tt\bold{s=27m.}$

So , The distance travelled by the car is 27m..