Question

The drive of a speedy car applied breaks which can produce an acceleration of 10ms² in a direction opposite to that of the montion. Calculate the distance it travels during 3s which is the time taken by the car before coming to a rest​

Answer 1

GivEn:

• Acceleration produced by the driver = 10 m/s²
• Time taken by the car before coming to a rest = 3 sec

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To find:

• Distance travelled by car.

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SoluTion:

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Here,

$\sf we\;have \begin{cases} & \text{v = 0 m/s } \\ & \text{a = - 10 m/ s^2 } \\ & \text{t = 3 s} \end{cases}$

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${\underline{\sf{\bigstar\;Using\;1st\;equation\;of\;motion\;:}}}$

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$\;\;\;\star\;{\boxed{\sf{\purple{v = u + at}}}}$

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$\;\;\;\;\;\small\sf \underline{Putting\;values}$

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$:\implies\sf 0 = u + (-10) \times 3$

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$:\implies\sf 0 = u - 30$

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$:\implies\bf \pink{u = 30\;m/s}$

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${\underline{\sf{\bigstar\;Again,\;Using\;2nd\;equation\;of\;motion\;:}}}$

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$\;\;\;\star\;{\boxed{\sf{\purple{s = ut + \dfrac{1}{2}at^2}}}}$

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$\;\;\;\;\;\small\sf \underline{Putting\;values}$

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$:\implies\sf s = 30 \times 3 + \dfrac{1}{2} \times -10 \times 3 \times 3$

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$:\implies\sf s = 90 + \dfrac{1}{ \cancel{2}} \times \cancel{-10} \times 9$

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$:\implies\sf s = 90 - 5 \times 9$

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$:\implies\sf s = 90 - 45$

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$:\implies{\underline{\boxed{\sf{\green{s = 45\;m}}}}}\;\bigstar$

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$\therefore$ Distance travelled by car before coming rest is 45 m.

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Additional Information:

★ There are three equations of motion:-

⠀⠀⠀✩ $\sf v = u + at$

⠀⠀⠀✩ $\sf s = ut + \dfrac{1}{2} at^2$

⠀⠀⠀✩ $\sf v^2 - u^2 = 2as$

⠀⠀ ━━━━━━━━━━━━━━━━━━━━━

$\;\;\star\;\sf Acceleration (a) = \dfrac{Final\; velocity (v) - Initial\; velocity (u)}{Time (t)}$

Answer 2

Given :

• Acceleration (a) = - 10 m/s²
• Time (t) = 3 seconds
• Final velocity (v) = 0 m/s

To find :

• Distance (s)

According to the question,

$\star \boxed{ \sf \red{v = u + at}}$

Now according to the formula we will calculate the initial velocity (u)

$\tt{ \: \: \rightarrow0 = u + ( - 10) \times 3 }$

$\sf{ \: \: \rightarrow0 = u + 30}$

$\sf{ \: \: \rightarrow\:u= 30-0}$

$\sf{ \: \: \rightarrow\:u= 30}\orange\bigstar$

Now we will calculate the distance, so the formula is,

$\star \boxed{ \sf \red{s = ut + \frac{1}{2} {at}^{2} }}$

So,according to the formula we will put the value

$\tt{ \: \: \rightarrow30 \times 3 + \frac{1}{2} \times ( - 10) \times {(3)}^{2} }$

$\tt{ \: \: \rightarrow90 + \frac{1}{ \cancel2} \times - \cancel{ 10 } {}^{ \: \: 5} \times 9 }$

$\tt{ \: \: \rightarrow90 - 5 \times 9}$

$\tt{ \: \: \rightarrow90 - 45}$

$\tt{ \: \: \rightarrow45\: m}\orange\bigstar$

So,the distance is 45 m...