
A car travelling with a speed $126$$kmph$ along a straight line comes to rest after travelling a distance $245$$m$. The time taken by the car to come to rest in seconds is?
(A) $11$
(B) $12$
(C) $16$
(D) $14$
Answer
552.6k+ views
Hint:Use the equations of kinematics to get the time required to come at rest. The body will retard with some deceleration in order to come to rest. The initial velocity and the distance travelled is given, use these data to find the desired quantity.
Formulae to be used:\[v = u + at\]$...\left( 1 \right)$
$s = ut + \dfrac{1}{2}a{t^2}$$...\left( 2 \right)$
Complete step by step solution:
In the above formulae, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time required to cover the distance $s$.
According to the given data, $v = 0$ as the car comes to rest,
$u = 126$$kmph$, here the initial velocity is given in $kmph$. Let us convert it into $m{s^{ - 1}}$
\[u = 126 \times \dfrac{{1000}}{{60 \times 60}} = 35\]$m{s^{ - 1}}$.
Now, let us substitute the value of acceleration from equation $\left( 1 \right)$in equation $\left( 2
\right)$,
$
v = u + at \\
v - u = at \\
\therefore a = \dfrac{{v - u}}{t} \\
$
$
s = ut + \dfrac{1}{2}\left( {\dfrac{{v - u}}{t}} \right){t^{^2}} \\
s = ut + \dfrac{1}{2}\left( {v - u} \right)t \\
s = ut + \dfrac{1}{2}vt - \dfrac{1}{2}ut \\
s = \dfrac{{ut}}{2} + \dfrac{{vt}}{2} \\
\therefore s = \left( {\dfrac{{u + v}}{2}} \right)t \\
$
From here we can obtain an expression to find the time taken by the car to come at rest.
$
s = \left( {\dfrac{{u + v}}{2}} \right)t \\
t = \dfrac{{2s}}{{u + v}} \\
$
Having \[u = 35\]$m{s^{ - 1}}$, distance covered as $s = 245$$m$and final velocity $v = 0$,
$t = \dfrac{{\left( 2 \right)\left( {245} \right)}}{{\left( {35 + 0} \right)}}$
$\therefore t = 14$$s$.
Therefore, the time taken by the car to come to rest in seconds is $14$$s$.
Option (D) is correct.
Note: Always keep in mind the units of the quantities given in the question and apply conversion accordingly, else you can get wrong answers even after substituting the correct values as given in the question, because they are not in the required units. Also remember that for conversion of velocity from $kmph$ to $m{s^{ - 1}}$ just multiply or divide the given velocity by $\dfrac{5}{{18}}$ as per requirements to save time.
Formulae to be used:\[v = u + at\]$...\left( 1 \right)$
$s = ut + \dfrac{1}{2}a{t^2}$$...\left( 2 \right)$
Complete step by step solution:
In the above formulae, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time required to cover the distance $s$.
According to the given data, $v = 0$ as the car comes to rest,
$u = 126$$kmph$, here the initial velocity is given in $kmph$. Let us convert it into $m{s^{ - 1}}$
\[u = 126 \times \dfrac{{1000}}{{60 \times 60}} = 35\]$m{s^{ - 1}}$.
Now, let us substitute the value of acceleration from equation $\left( 1 \right)$in equation $\left( 2
\right)$,
$
v = u + at \\
v - u = at \\
\therefore a = \dfrac{{v - u}}{t} \\
$
$
s = ut + \dfrac{1}{2}\left( {\dfrac{{v - u}}{t}} \right){t^{^2}} \\
s = ut + \dfrac{1}{2}\left( {v - u} \right)t \\
s = ut + \dfrac{1}{2}vt - \dfrac{1}{2}ut \\
s = \dfrac{{ut}}{2} + \dfrac{{vt}}{2} \\
\therefore s = \left( {\dfrac{{u + v}}{2}} \right)t \\
$
From here we can obtain an expression to find the time taken by the car to come at rest.
$
s = \left( {\dfrac{{u + v}}{2}} \right)t \\
t = \dfrac{{2s}}{{u + v}} \\
$
Having \[u = 35\]$m{s^{ - 1}}$, distance covered as $s = 245$$m$and final velocity $v = 0$,
$t = \dfrac{{\left( 2 \right)\left( {245} \right)}}{{\left( {35 + 0} \right)}}$
$\therefore t = 14$$s$.
Therefore, the time taken by the car to come to rest in seconds is $14$$s$.
Option (D) is correct.
Note: Always keep in mind the units of the quantities given in the question and apply conversion accordingly, else you can get wrong answers even after substituting the correct values as given in the question, because they are not in the required units. Also remember that for conversion of velocity from $kmph$ to $m{s^{ - 1}}$ just multiply or divide the given velocity by $\dfrac{5}{{18}}$ as per requirements to save time.
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