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A car of mass $1000 \mathrm{~kg}$ is moving at a speed of 30 $\mathrm{m} / \mathrm{s}$. Brakes are applied to bring the car to rest. If the net retarding force is $5000 \mathrm{~N}$, the car comes to stop after travelling $d \mathrm{~m}$ in $t \mathrm{~s}$. Then
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Verified Answer
The correct answer is:
$d=90, t=6$
$d=90, t=6$
Given: mass of car $m=1000 \mathrm{~kg}$
$$
\begin{aligned}
& u=30 \mathrm{~m} / \mathrm{s} \\
& v=0 \mathrm{~m} / \mathrm{s} \\
& \text { retarding force } f=5000 \mathrm{~N} \\
& \therefore \text { retardation, }-a=\frac{5000}{1000}=5 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
By equation, $v^2-u^2=2$ as
$$
\begin{aligned}
& 0-(30)^2=-2 \times 5 \times d \\
& \therefore d=\frac{900}{10}=90 \mathrm{~m}
\end{aligned}
$$
and $a=\frac{v-u}{t} \therefore t=\frac{v-u}{a}=\frac{0-30}{-5}=6 \mathrm{~s}$
$$
\begin{aligned}
& u=30 \mathrm{~m} / \mathrm{s} \\
& v=0 \mathrm{~m} / \mathrm{s} \\
& \text { retarding force } f=5000 \mathrm{~N} \\
& \therefore \text { retardation, }-a=\frac{5000}{1000}=5 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
By equation, $v^2-u^2=2$ as
$$
\begin{aligned}
& 0-(30)^2=-2 \times 5 \times d \\
& \therefore d=\frac{900}{10}=90 \mathrm{~m}
\end{aligned}
$$
and $a=\frac{v-u}{t} \therefore t=\frac{v-u}{a}=\frac{0-30}{-5}=6 \mathrm{~s}$
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