Question

A car moving along a straight highway at a speed of 144 km/hr is brought to a stop within a distance of 200 m.

(a) What is the retardation of the car (assumed uniform)?

(b) How long does it take for the car to stop?​

Answer 1

hi there here's your answer

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Answer 2

Provided that:

• Initial velocity = 144 km/h
• Final velocity = 0 m/s
• Distance = 200 m

Don't be confused! Final velocity cames as zero because the car stopped after travelling some distance with a velocity!

To calculate:

• Retardation
• Time taken

Solution:

• Retardation = -4 m/s sq.
• Time taken = 10 s

Using concepts:

• Third equation of motion
• Formula to convert kmph-mps
• First equation of motion

Using formulas:

• Third equation of motion is given by,

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

• First equation of motion is given by,

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

Required solution:

~ Firstly let us convert kmph-mps!

$:\implies \sf 144 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{144} \times \dfrac{5}{\cancel{{18}}} \\ \\ :\implies \sf 8 \times 5 \\ \\ :\implies \sf 40 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

~ Now let us find out the retardation!

$:\implies \sf v^2 \: - u^2 \: = 2as \\ \\ :\implies \sf (0)^{2} - (40)^{2} = 2(a)(200) \\ \\ :\implies \sf 0 - 1600 = (400a) \\ \\ :\implies \sf -1600 = 400 \times a \\ \\ :\implies \sf -1600/400 \: = a \\ \\ :\implies \sf -16/4 \: = a \\ \\ :\implies \sf -4 \: = a \\ \\ :\implies \sf a \: = -4 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -4 \: ms^{-2} \\ \\ :\implies \sf Retardation \: = -4 \: ms^{-2}$

~ Now let us calculate time!

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 0 = 40 + (-4)(t) \\ \\ :\implies \sf 0 = 40 + (-4t) \\ \\ :\implies \sf 0 = 40 - 4t \\ \\ :\implies \sf 0 - 40 = - 4t \\ \\ :\implies \sf -40 = - 4t \\ \\ :\implies \sf 40 = 4t \\ \\ :\implies \sf 40/4 \: = t \\ \\ :\implies \sf 10 \: = t \\ \\ :\implies \sf t \: = 10 \: seconds \\ \\ :\implies \sf Time \: = 10 \: seconds$

Additional information:

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