Question

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

Answer 1

Question.

The brakes applied to a car produce an acceleration of 6 m/s in the opposite

direction to the motion. If the car takes 2 s to stop after the application of

brakes, calculate the distance it travels during this time.

Solution.

Given:-

• Acceleration, a=-6 m/s^2.
• Final velocity, v=0 .
• Time taken to stop car =2 second.

Find the distance covered by car after brake.

Explanation:

We know that,

u=v-at

u=0-(-6)*2

u=12 m/s.

again,

s=ut+1/2a (t)^2.

s=12*2+1/2*(-6)*4

s=24+(-12)

s=24-12

s=12 m.

Therefore, Distance covered by car after brake is 12m.

Equation of motion.

__________________________________________

• v=u+1/2at.
• s=ut+1/2a(t)^2.
• v^2=u^2-2as.

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.