Question

driver taken 0.2 second to apply the brakes in p is moving at a speed of 54 km per hour and the break cause a ready retardation of 6 metre per second find the distance travelled by the car after he sees the need to put the break​

Answer 1

Given :-

Time taken by the car = 0.2 sec

Initial velocity of the car = 54 km/h

Retardation of the car = 6 m/s

To Find :-

The distance travelled by the car after he sees the need to put the break​.

Analysis :-

Here we are given with the time taken, initial velocity and the retardation of the car.

Firstly, find the distance covered by substituting the given values such that distance is equal to speed multiplied by time.

Then using the third equation of motion, you can easily find the distance travelled by the driver after he sees the need to put the break.

Solution :-

We know that,

• s = Distance
• u = Initial velocity
• a = Acceleration
• v = Final velocity
• t = Time

Using the formula,

$\underline{\boxed{\sf Distance=Speed \times Time}}$

Given that,

Speed (s) = 54 km/h = 15 m/s

Time (t) = 0.2 sec

Substituting their values,

⇒ d = s × t

⇒ d = 15 × 0.2

⇒ 3 m

Using the formula,

$\underline{\boxed{\sf Third \ equation \ of \ motion=v^2-u^2=2as}}$

Given that,

Final velocity (v) = 0 m/s

Initial velocity (u) = 15 m/s

Retardation (a) = 6 m/s

Substituting their values,

⇒ 0² - 15² = 2 × -6 × s

⇒ s = (-15 × 15) / 2 × -6

⇒ s = 225/12

⇒ s = 18.75 m

Total distance,

⇒ 18.75 + 3

⇒ 21.75 m

Therefore, the distance travelled by the car after he sees the need to put the break is 21.75 m.

Answer 2

Answer:

Question

driver taken 0.2 second to apply the brakes in p is moving at a speed of 54 km per hour and the break cause a ready retardation of 6 metre per second find the distance travelled by the car after he sees the need to put the break

Explanation:

Given in the question:-

• Time taken by the car = 0.2 sec
• Time taken by the car = 0.2 secInitial velocity of the car = 54 km/h
• Retardation of the car = 6 m/s

Find out in the question:-

• the distance travelled by the car after he sees the need to put the break

Here we have to use some short form of the letter

$</p><p>s = Distance \\ </p><p></p><p>u = Initial \: \ velocity \\ </p><p></p><p>a = Acceleration \\ </p><p></p><p>v = Final \: velocity \\ </p><p></p><p>t = Time$

Formula to be used here

$Distance=Speed×Time$

So,

Speed (s) = 54 km/h = 15 m/s

Time (t) = 0.2 sec

So finding distance travelled

⇒ distance = speed × time

⇒ d = 15 × 0.2

⇒ 3 m

Formula to be used again

$</p><p>\underline{\boxed{\sf Third \ equation \ of \ motion=v^2-u^2=2as}}$

So,

Final velocity (v) = 0 m/s

Initial velocity (u) = 15 m/s

Retardation (a) = 6 m/s

Putting the values

⇒ 0² - 15² = 2 × -6 × s

⇒ s = (-15 × 15) / 2 × -6

⇒ s = 225/12

⇒ s = 18.75 m

So, the total distance will be the sum of

18.75 + 3 = 21.5 m

•°• The required distance is

$21.5 \: metre \: or \: 21 \frac{1}{5}$

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