Question

The brakes apiled to a car produce an accerlaration of 6m/s^2 in the opposite direction to the motion.if the car takes 2s to stop after the application of the brakes ,calculate the distance it travels during this time

Answer 1

Answer:

12m

Explanation:

Given :

• Acceleration(a) => -6ms^2
• Time(t) => 2s
• Final velocity(v) => 0ms^-1

To Find :

• Distance it traveled

Solution :

We know that :-

$\Rightarrow$ v = u + at

Put their values

$\Rightarrow$ 0 = u + (-6ms^-2) × 2s

$\Rightarrow$ u =12ms^-1

So initial velocity is 12ms^-1

Therefore,Speed :-

$\sf{}s=ut+\dfrac{1}{2}at^2$

Put their values and solve

$\Rightarrow \sf{}(12ms^{-1})\times (2s)+\dfrac{1}{2}(-6ms^{-2})(2s)^{2}$

$\Rightarrow \sf{}24m -12m$

$\Rightarrow \sf{}12m$

Thus,the car will move 12m before it stops after the application.

Answer 2

Answer

• Distance , s = 12 m

Given

• The brakes applied to a car produce an acceleration of 6 m/s² in the opposite direction to the motion.if the car takes 2 s to stop after the application of the brakes

To Find

• Distance travelled

Concept Used

$\boxed{\begin{minipage}{4cm} \underbrace{ \bf Equations\ of\ motion\ :}\\\\\rm \bigstar \;\; v=u+at \\\\\rm \bigstar \;\; s=ut+\dfrac{1}{2}at^2\\\\\rm \bigstar \;\; v^2-u^2=2as\\\\\rm \bigstar \;\; S_n=u+\dfrac{a}{2}[2n-1] \end{minipage}}$

Solution

Acceleration , a = - 6 m/s²

[ ∵ acts opposite direction to the motion ]

Final velocity , v = 0 m/s

[ ∵ Finally stops after applying brakes ]

Time , t = 2 s

Apply 1 st equation of motion .

⇒ v = u + at

⇒ (0) = u + (-6)(2)

⇒ 0 = u - 12

⇒ 12 = u

⇒ u = 12 m/s

So , Initial velocity , u = 12 m/s

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Apply 3 rd equation of motion .

⇒ v² - u² = 2as

⇒ (0)² - (12)² = 2(-6)s

⇒ 0 - 144 = -12s

⇒ - 144 = -12s

⇒ 144 = 12s

⇒ 12s = 144

⇒ 12s = 12 × 12

⇒ s = 12 m

So , Car  travels 12 m during the time 2 seconds .

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