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4-1.Newton's Laws of Motion
hard
A car of mass $1000\,kg$ is moving at a speed of $30\,m/s.$ Brakes are applied to bring the car to rest. If the net retarding force is $5000\,N,$ the car comes to stop after travelling $d\,m$ in $t\,s.$ Then
A
$d = 150,\,t = 5$
B
$d= 120,\,t = 8$
C
$d = 180,\,t = 6$
D
$d = 90,\,t = 6$
(AIEEE-2012)
Solution
$\begin{array}{l}
Giverr.\,mass\,of\,car\,m = \,1000\,kg\\
u = 30\,m/s\\
v = 0\,m/s\\
retarding\,force\,f = 5000\,N\\
\therefore \,retardation, – a = \frac{{5000}}{{1000}} = 5m/{s^2}\\
By\,equartion,\,{v^2} – {u^2} = 2as\\
0 – {\left( {30} \right)^2} = – 2 \times 5 \times d\\
\therefore d = \frac{{900}}{{10}} = 90\,m\\
and\,a = \frac{{v – u}}{t}\therefore t = \frac{{v – u}}{a} = \frac{{0 – 30}}{{ – 5}} = 6s
\end{array}$
Standard 11
Physics
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