Question

Car is travelling at a speed of 90 km/h.Brakes are applied so as to produce a uniform acceleration of -0.5 m/s² .Find how far the car will go before it is brought to rest?

please explain ful answer​

Answer 1

Heya ! here's ur solution :D

$\large \frak{given} \begin{cases} \sf initial \:velocity = 90 \: km \: {hr}^{ - 1} \\ \sf uniform \: accelaration = - 0.5m {s}^{ - 2} \\ \sf final \: velocity \: = 0 \: km \: {hr}^{ - 1} \end{cases}$

• see, I assumed that the final velocity of the car is 0 because it is going under retardation and is being brought to rest.

• We are asked to calculate the distance traveled by this vehicle before coming to rest.

• First of all convert the Given value of initial velocity in m/s

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bullet \: \sf 1km \: {hr}^{ - 1} = \dfrac{5}{18} m {s}^{ - 1}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bullet \: \sf 90km \: {hr}^{ - 1} = \cancel{90} \times \dfrac{5}{ \cancel{18}} = 25m {s}^{ - 1}$

Now, we will be using 3rd equation of motion in order to find the required value

We know that :-

$\longrightarrow \large \underline{ \boxed{ \pink{{ \frak{ { \sf v}^{2} - {u}^{2} = 2as}}}}} \bigstar$

• Here, v, u, a and s represents final velocity, initial velocity, Acceleration and distance respectively.

$\: \: \: \: \: \: \: \: \dag \: \underline \frak{substituting \: the \: values \: in \: equation : - }$

$\: \: \: \: \: \: \: \: : \implies \sf {(0)}^{2} - {(25)}^{2} =( 2 ) \times ( - 0.5 ) \times s$

$\: \: \: \: \: \: \: \: : \implies \sf 0 - 625 = - s$

$\: \: \: \: \: \: \: \: : \implies \sf \cancel{- }625 = \cancel{-} s$

$\: \: \: \: \: \: \: \: : \implies {\boxed{ \pink{{ \frak{s = 625 \: m}}}}} \bigstar$

$\therefore \underline {\sf the \: car \: travelled \: a \: distance \: of \: 625 \: m \: before \: coming \: to \: rest }$

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Hope it will help you !