A driver applies the brakes on seeing the red traffic signal 400 m ahead. At the time of applying brakes,…
A driver applies the brakes on seeing the red traffic signal 400 m ahead. At the time of applying brakes, the vehicle was moving with $15 \mathrm{~m} / \mathrm{s}$ and retarding at $0.3 \mathrm{~m} / \mathrm{s}^2$. The distance of the vehicle from the traffic light one minute after application of brakes is
375 m
360 m
40 m
25 m
Solution
After applying the brakes, the vehicle stops after time t ,
$\mathrm{t}=\left(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{a}}\right)=\left(\frac{0-15}{-0.3}\right)=50 \text { seconds }$
i.e. vehicle stops before one minute.
$\Rightarrow$ Displacement will only occur for 50 seconds.
$\therefore \quad$ Displacement is given by $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2$
$\therefore \quad \mathrm{s}=15 \times 50+\frac{1}{2} \times(-0.3) \times(50)^2$
$\ldots .(\because \mathrm{a}$ is the retardation in vehicle)
$\mathrm{s}=375 \mathrm{~m}$ Distance from traffic light $=400-375=25 \mathrm{~m}$