Question

A car moving at a speed of 30 km per hour is brought to a halt in 8 metres by applying brakes. If the same car is moving at 60 Km per hour , it can be brought to halt with same braking power in ?



(answer- 32 metres)​

Answer 1

Answer:

32m

Explanation:

see the above page and you'll get to know the answer perfectly with explanation..

hope it helps..

Answer 2

★ Concept :-

Here the concept of Newton's Equation of Motions has been used . We see that in first case the initial velocity of car is different than second. But for both the cases final velocity is same as zero because cars are being brought to halt. So in first case we can firstly find the acceleration of the care. Then using that acceleration because same car is moving, we can find the distance travelled by applying brakes to come to halt.

Let's do it !!

______________________________________

★ Formula Used :-

$\;\boxed{\sf{\pink{v^{2}\:-\:u^{2}\:=\:\bf{2as}}}}$

This is Newton's Third Equation of Motion.

______________________________________

★ Solution :-

• For First Case ::

• Initial velocity = u = 30 Km/hr = 30 × 5/18 m/sec = 8.33 m/sec

(units must be same for calculation)

• Distance Travelled = s = 8 metres

• Let the acceleration be a

• Final Velocity = v = 0

We know that,

$\;\sf{\rightarrow\;\;v^{2}\:-\:u^{2}\:=\:\bf{2as}}$

By applying values here, we get

$\;\sf{\rightarrow\;\;(0)^{2}\:-\:(8.33)^{2}\:=\:\bf{2(a)(8)}}$

$\;\sf{\rightarrow\;\;-\:69.389\:=\:\bf{16a}}$

$\;\sf{\rightarrow\;\;a\:=\:\bf{\dfrac{-69.388}{16}}}$

$\;\sf{\rightarrow\;\;a\:=\:\bf{\red{-\:4.33\;\:m/sec²}}}$

Here -ve sign shows the retardation as body is brought to rest by applying breaks.

• For Second Case ::

• Initial velocity = u = 60 Km/h = 60 × 5/18 m/sec = 16.67 m/sec

(units must be same for calculation)

• Acceleration = a = - 4.33 m/sec²

(since body is brought to rest, acceleration is negative)

• Distance travelled = s

• Final Velocity = v = 0

By using the formula of Newton's Third Equation of Motion, we get

$\;\sf{\rightarrow\;\;v^{2}\:-\:u^{2}\:=\:\bf{2as}}$

By applying values , we get

$\;\sf{\rightarrow\;\;(0)^{2}\:-\:(16.67)^{2}\:=\:\bf{2(-4.33)(s)}}$

$\;\sf{\rightarrow\;\;-\:277.89\:=\:\bf{-8.66s}}$

$\;\sf{\rightarrow\;\;s\:=\:\bf{\dfrac{-277.89}{-8.66}}}$

$\;\sf{\rightarrow\;\;s\:=\:\bf{\green{32\;\:m}}}$

(since 32.088 ≈ 32)

This is the required answer.

Body will be brought to rest after coverin 32 m.

$\;\underline{\boxed{\tt{Required\;\:Distance\;=\;\bf{\purple{32\;\:m}}}}}$

________________________________

★ More to know :-

$\;\tt{\leadsto\;\:v\:-\:u\:=\:at}$

$\;\tt{\leadsto\;\:s\;=\;ut\;+\;\dfrac{1}{2}at^{2}}$

$\;\tt{\leadsto\;\; Velocity\;=\;\dfrac{Displacement}{Time}}$

$\;\tt{\leadsto\;\; Acceleration\;=\;\dfrac{Change\:in\:Velocity}{Time}}$

$\;\tt{\leadsto\;\;v\;=\;\sqrt{2as}}$

If u = 0