Question

The brakes applied to a car produce an acceleration of 6 m/s² in the opposite direction to the motion If the car takes 2 seconds to stop after the application of brakes, calculate the distance it travels during this time.

Answer 1

Hello
Formula =v=u + ats = ut + 1/2at²
                         Final Velocity = v = 0 {it is at rest}                         Initial Velocity = Unkown = u                         s = distance                         a = acceleration { -ve retardation}v= u + at0 = u +(-6) x 2u= 12m/s
s= ut+1/2at²s=36 meters
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Answer 2

Given:-

• Acceleration ,a = 6m/s² (acting in opposite direction)

• Final velocity ,v = 0m/s

• Time taken ,t = 2s

To Find:-

• Distance travelled ,s

Solution:-

According to the Question$,$

Firstly we calculate the initial velocity of the car .

By using Kinematics Equation

• v = u + at

where,

• v is the final velocity

• a is the acceleration

• u is the initial velocity

• t is the time taken

Putting all the given value s we get

➺ 0 = u + (-6) × 2

➺ 0 = u -12

➺ -u = -12m/s

➺ u = 12m/s

So, the initial velocity of the car was 12m/s.

Now, calculating the distance covered by car . Again by using Kinematics Equation

• v² = u² + 2as

Substitute the value we get

➺ 0² = 12² + 2×(-6) × s

➺ 0 = 144 + (-12) × s

➺ -144 = -12× s

➺ 144 = 12×s

➺ s = 144/12

➺ s = 12m

• Hence, the distance covered by the car during this time is 12 metres.