Question

A car moves 40km per hour It is stopped appyling breaks Which produces a uniform Acceleration of 0.6 meter per sec How much distance will the vehicle move before stopping

Answer 1

Answer :

The distance covered by the car is 102.88m

Given :

• A car is moving at 40km/h .
• It stopped by applying brakes at constant acceleration of 0.6m/s²

To Find :

• The distance the car travelled before it stopped.

Formula to be used :

If v is final velocity ,u is initial velocity , a is the acceleration and s is the distance covered then according to equations of motion :

$\sf \bullet \: \: v^{2} - u^{2} = 2as$

Solution :

Given ,

$\sf u = 40km{h}^{-1} \\\\ \sf \implies u = 40\times \dfrac{5}{18}ms^{-1}\\\\ \sf \implies u = \dfrac{100}{9}ms^{-1}$

v = 0m/s

a = -0.6m/s² [ since the car stopped at the end so the acceleration is in opposite direction and taken as negative]

Now using equations of motion :

$\implies\sf 0^{2} -(\dfrac{100}{9})^{2} = 2\times (-0.6) \times s \\\\ \sf \implies -1.2s =- \dfrac{100 \times 100}{9\times 9} \\\\ \sf \implies s = \dfrac{100 \times 100}{9 \times 9 \times 1.2} \\\\ \sf \implies s = 102.88m$

Answer 2

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$\huge\tt {GIVEN:}$

• A car moves 40km per hour It is stopped applying breaks
• Which produces a uniform Acceleration of 0.6 meter per sec

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$\huge\tt {TO~FIND:}$

• How much distance will the vehicle move before stopping

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$\huge\tt {SOLUTION:}$

If Final velocity is v, Initial Velocity would be u, acceleration is A, and Distance is S , then the equation would be-

• v² - u² = 2as

Then,

↪U = 40 × 5/18 ms -¹

↪U = 100/9 ms -¹

↪U = 0 m/sec

a = -0.6 m/s

Now, the equations of motions would be,

↪0² - (100/9)² = 2 × (-0.6) × s

↪-1.2s = 1000/81

↪s = 1000/81 × 1.2

↪s = 102.88 m

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