Question

An automobile travelling with a speed of 60 km /hr can break to stop within a distance of 20 m. if the car is going twice as fast i.e,120km /hr ,the stopping distance will be.​

Answer 1

GIVEN :-

• An automobile travelling with a speed of 60 km/hr can break to stop within a distance of 20 m.

TO FIND :-

• Stopping distance , if the car is going twice as fast i.e, 120km/hr.

SOLUTION :-

We have ,

♦ Initial velocity (u) =60km/hr

ㅤㅤㅤㅤㅤㅤㅤㅤ= 60 × (5/18) m/s

ㅤㅤㅤㅤㅤㅤㅤㅤ= 16.66m/s

♦ Final velocity (v) = 0m/s (as car is stopped)

♦ Stopping distance (s) = 20m

Now , by 3rd Kinematical equation,

★ v² = u² + 2as

Putting values,

→ 0² = 16.66² + 2(a)(20)

→ 0 = 277.55 + 40a

→ -277.55 = 40a

→ a = -277.55/40

→ a = -6.93m/s²

Hence , acceleration of the car is -6.93m/s².

Now , In another case ,

♦ Initial velocity(u) = 120km/hr

ㅤㅤㅤㅤ ㅤㅤ ㅤㅤ= 120 × (5/18) m/s

ㅤㅤㅤ ㅤㅤ ㅤㅤㅤ= 33.33m/s

♦ Final velocity (v) = 0m/s (car is stopped)

♦ a = -6.93m/s²

Again , by 3rd Kinematical equation,

★ v² = u² + 2as

Putting values,

→ 0² = 33.33² + 2(-6.93)(s)

→ 0 = 1110.88 - 13.86s

→ -13.86s = -1110.88

→ s = -1110.88/-13.86

→ s = 80.15m ≈ 80m

Hence , if the car is going twice as fast i.e,120km/hr, the stopping distance will be 80m.

Answer 2

Given :

• An automobile travelling with a speed of 60 km /hr
• Stop within a distance of 20 m.
• Car is going twice as fast i.e,120km /hr

To find :

• Stopping distance will be .

Solution :

$\sf \;\;\bf{\;\;\red{Let \: us \: see \: the \: 1st \: equation . }} \:$

$\sf0 = ( \frac{50}{3} ) {}^{2} - 2a \times 20$

$\sf \;\;\bf{\;\;\blue{Let \: us \: see \: the \: 2nd \: equation . }} \:$

$\sf0 = ( \frac{100}{3} ) {}^{2} - 2ax$

$\;\;\bf{\;\;\green{x = 80 \: m}}$

$\;\;\bf{\;\;\blue{Hence , The}} \bf{ \: \red{answer \: is \: }}\bf{\green{x = 80 \: m \: }}$