Question

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

Answer 1

Answer:

a= -6m/s^2. ( acts in a opposite direction)

Now,

we know-

a= -6m/s^2

u=0

t=2seconds

so,we have to find s=?

therefore,by using second equation of motion,

s=ut+1/2at^2

s=(0×2)+1/2×-6×(2)^2

s=0+1/2×6×4

s=12m

therefore,distance=12m

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.