Question

a car is moving with speed of 20m/s after applying the brakes if the maximum deceleration produced in 10m/s^2 then find the distance covered by the car before coming to rest? ​

Answer 1

Given :

• Initial velocity, u = 20 m/s

• Final velocity, v = 0 m/s (Brakes applied)

• Acceleration, a = - 10 m/s²

To find :

• Distance covered by the car, s

According to the question,

➞ v = u + at

Where,

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

➞ 0 = 20 + (-10) × t

➞ 0 - 20 = - 10t

➞ - 20 = - 10t

➞ 20 ÷ 10 = t

➞ 2 = t

So,the time taken is 2 seconds.

Now,

➞ s = ut + ½ at²

Where,

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

➞ Substituting the values,

➞ s = 20 × 2 + ½ × (-10) × 2 × 2

➞ s = 40 + (-10) × 2

➞ s = 40 + (-20)

➞ s = 40 - 20

➞ s = 20

So,the distance covered by the car is 20 meters.

Answer 2

$\huge{\boxed{\rm{Question}}}$

A car is moving with speed of 20m/s after applying the brakes if the maximum acceleration produced in 10m/s² then find the distance covered by the car before coming to rest ?

$\huge{\boxed{\rm{Answer}}}$

$\large{\boxed{\boxed{\sf{Given \: that}}}}$

• Initial velocity = 20 m/s

• Final velocity = 0 m/s

• Acceleration = 10 m/s²

$\small{\boxed{\boxed{\sf{R \: u \: confused \: due \: to \: 0 m/s \: as \: final \: velocity}}}}$

$\small{\boxed{\boxed{\sf{Yes \: we \: r \: little \: bit \: confused}}}}$

$\small{\boxed{\boxed{\sf{No \: need \: to \: worry \: let's \: see \: why \: it's \: like \: this}}}}$

✞ The final velocity is 0 m/s because the car is at rest. And in question it's already said that it applying brake that time. That's why the final velocity be 0 m/s.

Nᴏᴡ ᴄᴏᴍɪɴɢ ᴛᴏ ᴛʜᴇ ǫᴜᴇsᴛɪᴏɴ

$\large{\boxed{\boxed{\sf{Given \: that}}}}$

• Distance.

$\large{\boxed{\boxed{\sf{Solution}}}}$

• Distance = 20 metres.

$\large{\boxed{\boxed{\sf{We \: also \: write \: these \: as}}}}$

• Final velocity as v.

• Initial velocity as u.

• Acceleration as a.

• Time as t.

• Distance as s.

$\large{\boxed{\boxed{\sf{What \: does \: the \: question \: says}}}}$

$\large{\boxed{\boxed{\sf{Let's \: understand \: the \: concept \: 1st}}}}$

• This question says that there is a car which is moving with speed of 20m/s after applying the brakes if the maximum acceleration produced in 10m/s² then it ask us to find the distance covered by the car before coming to rest.

$\large{\boxed{\boxed{\sf{How \: to \: do \: this \: question}}}}$

$\large{\boxed{\boxed{\sf{Let's \: see \: the \: procedure \: now}}}}$

• To solve this question 1stly we have to learn the formula of Newton's law of motion and we already know them. So, now coming to question 1stly we have to use the formula of Newton's first law of motion and it's v = u + at. Now we have to put the values according to this formula afterthat we get 2 seconds as the result. Means time taken is 2 seconds. After that using the formula of Newton's 2nd law of motion and it's s = ut + ½ at² afterthat putting the values according to this formula we get 20 m as final result. Hence, it's total distance taken by car.

$\large{\boxed{\boxed{\sf{Full \: solution}}}}$

We have to use this formula to solve this question ➝ v = u + at.

Where,

• v means final velocity.

• u means initial velocity

• a means acceleration

• t as time.

Putting the values we get the following results ➝

➥ v = u + at.

➥ 0 = 20 + (-10) × t

➥ 0 - 20 = -10t

➥ -20 = -10t { Cancelling - by - = + }

➥ 20 / 10 t

➥ 2/1 t

➥ 2t

Hence, the taken time ( t ) = 2 seconds.

We have to use this formula to solve this question ➝ s = ut + ½ at²

Where,

• s means distance

• u means initial velocity

• t means time

• a means acceleration

• ² means square.

Putting the values we get the following results ➝

➥ s = ut + ½ at²

➥ s = 20 × 2 + ½ × (-10) × 2 × 2

➥ s = 40 + (-10) × 2

➥ s = 40 + (-20)

➥ s = 40-20

➥ s = 20

Hence, 20 m is the distance.