Question

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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 2 s to stop after the application of brakes, calculate the distance it travels during this time.


Note:-
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Answer 1

Answer:

• 12m is the required answer .

Explanation:

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.

Answer 2

Given :-

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 2 s to stop after the application of brakes,

To Find :-

Distance covered

Solution :-

We know that

v = u + at

Where

v = final velocity

u = initial velocity

a = acceleration

t = time

0 = u + (-6)(2)

0 - u = -12

-u = -12

u = 12 m/s

Now,

s = ut + 1/2 at²

Where

s = distance

u = initial velocity

a = acceleration

t = time

s = 12 × 2 + 1/2 × -6 × (2)²

s = 24 + 1/2 × -6 × 4

s = 24 + (-12)

s = 24 - 12

s = 12 m