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Consider a vechicle moving with a velocity $54 \mathrm{~km} / \mathrm{h}$. At a distance of $400 \mathrm{~m}$, from the traffic light brakes are applied. The acceleration of the vehicle, after the application of brakes is $-0.3 \mathrm{~m} / \mathrm{s}^2$. The vehicle's position relative to the traffic light is
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The correct answer is:
$25 \mathrm{~m}$
Given, initial velocity,
$$
\begin{aligned}
& u=54 \mathrm{~km} / \mathrm{h}=54 \times \frac{5}{18} \mathrm{~m} / \mathrm{s} \\
& u=15 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Distance of signal from vehicle, $d=400 \mathrm{~m}$ acceleration, $a=-0.3 \mathrm{~m} / \mathrm{s}^2$
Now, distance covered by vehicle, when it stops is given by $v^2=u^2+2 a s$
$$
0=(15)^2-2 \times 0.3 \times s \Rightarrow s=\frac{225}{0.6}=375 \mathrm{~m}
$$
Relative position $=400-375=25 \mathrm{~m}$
$$
\begin{aligned}
& u=54 \mathrm{~km} / \mathrm{h}=54 \times \frac{5}{18} \mathrm{~m} / \mathrm{s} \\
& u=15 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Distance of signal from vehicle, $d=400 \mathrm{~m}$ acceleration, $a=-0.3 \mathrm{~m} / \mathrm{s}^2$
Now, distance covered by vehicle, when it stops is given by $v^2=u^2+2 a s$
$$
0=(15)^2-2 \times 0.3 \times s \Rightarrow s=\frac{225}{0.6}=375 \mathrm{~m}
$$
Relative position $=400-375=25 \mathrm{~m}$
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