Question

8.A car is moving with a velocity of 18 m/s. It is brought to rest by applying the brakes which produce a retardation of 0.5 m/s^2. The time taken by the car to come to rest is
(1 Point)

20 s

40 s

50 s

100 s​

Answer 1

➤ Answer

• Time taken by the car to come to rest = 36 seconds.

➤ Given

• A car is moving with a velocity of 18 m/s. It is brought to rest by applying the brakes which produce a retardation of 0.5 m/s².

➤ To Find

• Time taken by the car to come to rest.

➤ Step By Step Explanation

Given that ⤵

• Initial velocity = 18m/s
• Final velocity = 0m/s
• Acceleration = -0.5 m/s²

We need to calculate the time taken by the car to come to rest.

So let's do it !!

➠ By using first equation of motion

• v = u + at

➠ By substituting the values

$\longmapsto\tt0 = 18 + ( - 0.5) \times t \\ \\\longmapsto\tt (-18) = (- 0.5t) \\ \\\longmapsto\tt \cancel\cfrac{ - 18}{ - 0.5} = t \\ \\\longmapsto\tt 36 \: sec = t$

Therefore time taken by the car to come to rest = 36 seconds.

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➤ Equations of motion

There are mainly three equations of motion. They are as follows -

• v = u + at
• s = ut + ½at²
• v² - u² = 2as

Where, u = initial velocity , v = final velocity, t = time, a = acceleration, and s = distance travelled.

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Answer 2

Provided that:

↪️ Initial velocity = 18 m/s

↪️ Final velocity = 0 m/s

↪️ Retardation = -0.5 m/s²

Don't be confused!

• Final velocity cames as zero because it is given that break applied to a car to brought it as rest after moving!

• I write -0.5 m/s² except of +0.5 m/s² because it is provided that it retards, and retardation is always negative.

To calculate:

• The time taken

Solution:

• The time taken = 36 seconds

Using concept:

• First equation of motion

Using formula:

• ${\small{\underline{\boxed{\sf{v \: = u \: + at}}}}}$

Where, u denotes the initial velocity, t denotes time taken, a denotes acceleration and v denotes final velocity.

Required solution:

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 0 = 18 + (-0.5)(t) \\ \\ :\implies \sf 0 = 18 + (-0.5t) \\ \\ :\implies \sf 0 = 18 - 0.5t \\ \\ :\implies \sf 0 - 18 = -0.5t \\ \\ :\implies \sf -18 = -0.5t \\ \\ :\implies \sf 18 = 0.5t \\ \\ :\implies \sf \dfrac{18}{0.5} \: = t \\ \\ :\implies \sf \dfrac{180}{5} \: = t \\ \\ :\implies \sf \cancel{\dfrac{180}{5}} \: = t \\ \\ :\implies \sf 36 \: seconds \\ \\ :\implies \sf Time \: = 36 \: seconds$

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• Dear user, according to correct calculations the correct answer is 36 seconds except of the provided answers. Thanks for understanding!

Additional information:

$\begin{gathered}\boxed{\begin{array}{c|cc}\bf Speed&\bf Velocity\\\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}&\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}\\\sf The \: distance \: travelled \: by &\sf The \: distance \: travelled \: by \\ \sf \: a \: body \: per \: unit \: time&\sf \: a \: body \: per \: unit \: time \\ &\sf in \: a \: given \: direction \\\\\sf It \: is \: scalar \: quantity. &\sf It \: is \: vector \: quantity \\\\\sf It \: is \: positive \: always &\sf It \: can \: be \: \pm \: \& \: 0 \: too \\\\\sf Speed \: = \dfrac{Distance}{Time} &\sf Velocity \: = \dfrac{Displacement}{Time} \end{array}}\end{gathered}$

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• Dear web users, you can see step of cancelling from attachment 1st and 2nd.