MHT CET · Physics · Motion In One Dimension
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
- A 375 m
- B 360 m
- C 40 m
- D 25 m
Answer & Solution
Correct Answer
(D) 25 m
Step-by-step Solution
Detailed explanation
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 \(=400-375=25 \mathrm{~m}\)
\(\mathrm{t}=\left(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{a}}\right)=\left(\frac{0-15}{-0.3}\right)=50 \text { seconds }\)
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 \(=400-375=25 \mathrm{~m}\)
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