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A driver applies the brakes on seeing the red traffic single $400 \mathrm{~m}$ ahead. At the time of applying the brakes, the vehicle was moving with $15 \mathrm{~m} / \mathrm{s}$ and retarding at $0.3 \mathrm{~m} / \mathrm{s}^2$. The distance covered by the vehicle from the traffic light 1 minute after the application of brakes is
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Verified Answer
The correct answer is:
40 m
$$
\begin{aligned}
& \mathrm{u}=15 \mathrm{~m} / \mathrm{s}, \mathrm{a}=-0.3 \mathrm{~m} / \mathrm{s}^2, \mathrm{t}=60 \mathrm{~s} \\
& \mathrm{~s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 \\
& =15 \times 60+\frac{1}{2} \times(-0.3) \times(60)^2 \\
& =900-540=360 \mathrm{~m}
\end{aligned}
$$
Distance from traffic light $=400-360=40 \mathrm{~m}$
\begin{aligned}
& \mathrm{u}=15 \mathrm{~m} / \mathrm{s}, \mathrm{a}=-0.3 \mathrm{~m} / \mathrm{s}^2, \mathrm{t}=60 \mathrm{~s} \\
& \mathrm{~s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 \\
& =15 \times 60+\frac{1}{2} \times(-0.3) \times(60)^2 \\
& =900-540=360 \mathrm{~m}
\end{aligned}
$$
Distance from traffic light $=400-360=40 \mathrm{~m}$
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