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A car is moving with a speed of $54 \mathrm{~km} / \mathrm{h}$. If after $3 \mathrm{~s}$, the driver applies brakes and it stops, then how much distance is covered by the car before coming to rest?
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Verified Answer
The correct answer is:
$22.5 \mathrm{~m}$
Given, speed of car, $u=54 \mathrm{~km} / \mathrm{h}=54 \times \frac{5}{18}$
$$
=15 \mathrm{~m} / \mathrm{s}
$$
Since, car stops in $3 \mathrm{~s}$ on applying brakes, so acceleration can be calculated from first equation of motion,
$$
\begin{aligned}
v & =u+a t \Rightarrow 0=15+a(3) \\
\Rightarrow \quad & a=-\frac{15}{3}=-5 \mathrm{~ms}^{-2}
\end{aligned}
$$
Using second equation of motion,
$$
\begin{aligned}
s & =u t+\frac{1}{2} a t^2 \\
& =15 \times 3+\frac{1}{2} \times(-5) \times(3)^2 \\
& =45-22.5=22.5 \mathrm{~m} .
\end{aligned}
$$
$$
=15 \mathrm{~m} / \mathrm{s}
$$
Since, car stops in $3 \mathrm{~s}$ on applying brakes, so acceleration can be calculated from first equation of motion,
$$
\begin{aligned}
v & =u+a t \Rightarrow 0=15+a(3) \\
\Rightarrow \quad & a=-\frac{15}{3}=-5 \mathrm{~ms}^{-2}
\end{aligned}
$$
Using second equation of motion,
$$
\begin{aligned}
s & =u t+\frac{1}{2} a t^2 \\
& =15 \times 3+\frac{1}{2} \times(-5) \times(3)^2 \\
& =45-22.5=22.5 \mathrm{~m} .
\end{aligned}
$$
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