Physics, asked by gramagranny, 9 months ago

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

Answers

Answered by Anonymous
38

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.


Anonymous: Nice Explanation :)
Answered by BrainlyIAS
21

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

__________________________

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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