Question

VA car starting from rest acquires a velocity of
36 km h-' in 5 s. Calculate:
(a) its acceleration.
[Ans. 2 ms)
(b) the distance covered by it.
[Ans. 25 m​

Answer 1

Correct question: A car starting from rest acquires a velocity of 36 km/h in 5 seconds. Calculate:

• (a) Its acceleration.
• (b) The distance covered by it.

Provided that:

• Final velocity = 36 km/h
• Initial velocity = 0 m/s
• Time taken = 5 seconds

Don't be confused! Initial velocity cames as zero because the car starts from rest.

To calculate:

• Its acceleration.
• The distance covered by it.

Solution:

• Its acceleration = 2 m/s sq.
• The distance covered by it = 25 m

Using concepts:

• Formula to convert km/h into m/s

• Acceleration is given by

• ${\small{\underline{\boxed{\sf{a \: = \dfrac{v-u}{t}}}}}}$

• Newton's thrid equation of motion is given by the mentioned formula:

• ${\small{\underline{\boxed{\sf{2as \: = v^2 \: - u^2}}}}}$

Where, v denotes final velocity, u denotes initial velocity, t denotes time taken, a denotes acceleration, s denotes displacement or distance or height.

Required solution:

~ Firstly let us convert km/h into m/s by using suitable formula!

Converting 36 km/h into m/s

$:\implies \sf 36 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{36} \times \dfrac{5}{\cancel{{18}}} \\ \\ :\implies \sf 2 \times 5 \\ \\ :\implies \sf 10 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Dear web users, you can see step of cancelling from the attachment 1st

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~ Now let's find out the acceleration!

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{10-0}{5} \\ \\ :\implies \sf a \: = \dfrac{10}{5} \\ \\ :\implies \sf a \: = \cancel{\dfrac{10}{5}} \\ \\ :\implies \sf a \: = 2 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 2 \: ms^{-2} \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

Dear web users, you can see step of cancelling from the attachment 2nd

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~ Now let's find out the distance covered by the car by using third equation of motion, let's do it!

$:\implies \sf 2as \: = v^2 \: - u^2 \\ \\ :\implies \sf 2(2)(s) \: = (10)^{2} - (0)^{2} \\ \\ :\implies \sf 2(2)(s) \: = 100 - 0 \\ \\ :\implies \sf 2(2)(s) \: = 100 \\ \\ :\implies \sf 2 \times 2 \times s \: = 100 \\ \\ :\implies \sf 4 \times s \: = 100 \\ \\ :\implies \sf 4s \: = 100 \\ \\ :\implies \sf s \: = \dfrac{100}{4} \\ \\ :\implies \sf s \: = \cancel{\dfrac{100}{4}} \\ \\ :\implies \sf s \: = 25 \: m \\ \\ :\implies \sf Distance \: = 25 \: metres \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

Dear web users, you can see step of cancelling from the attachment 3rd

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About acceleration:

$\begin{gathered}\boxed{\begin{array}{c}\\ \bf What \: is \: acceleration? \\ \\ \sf The \: rate \: of \: change \: of \: velocity \: of \: an \\ \sf object \: with \: respect \: to \: time \\ \sf is \: known \: as \: acceleration. \\ \\ \sf \star \: Negative \: acceleration is \: known \: as \: deacceleration. \\ \sf \star \: Deacceleration \: is \: known \: as \: retardation. \\ \sf \star \: It's \: SI \: unit \: is \: ms^{-2} \: or \: m/s^2 \\ \sf \star \: It \: may \: be \: \pm ve \: or \: 0 \: too \\ \sf \star \: It \: is \: a \: vector \: quantity \\ \\ \bf Conditions \: of \pm ve \: or \: 0 \: acceleration \\ \\ \sf \odot \: Positive \: acceleration: \: \sf When \: \bf{u} \: \sf is \: lower \: than \: \bf{v} \\ \sf \odot \: Negative \: acceleration: \: \sf When \: \bf{v} \: \sf is \: lower \: than \: \bf{u} \\ \sf \odot \: Zero \: acceleration: \: \sf When \: \bf{v} \: \sf and \: \bf{u} \: \sf are \: equal \end{array}}\end{gathered}$

Difference between distance and displacement:

$\begin{gathered}\boxed{\begin{array}{c|cc}\bf Distance&\bf Displacement\\\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}&\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}\\\sf Path \: of \: length \: from \: which &\sf The \: shortest \: distance \: between \\ \sf \: object \: is \: travelling \: called \: distance. &\sf \: the \: initial \: point \: \& \: final \\ &\sf point \: is \: called \: displacement. \\\\\sf It \: is \: scalar \: quantity. &\sf It \: is \: vector \: quantity \\\\\sf It \: is \: positive \: always &\sf It \: can \: be \: \pm \: \& \: 0 \: too \end{array}}\end{gathered}$

Attachment 4th & 5th respectively!