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Chapter 2: Problem 202

An automobile travelling at a speed of \(60 \mathrm{~km} / \mathrm{h}\) can brake to stop within a distance of \(20 \mathrm{~m}\). If the car is going twice as fast, i.e. \(120 \mathrm{~km} / \mathrm{h}\), the stopping distance will be (A) \(60 \mathrm{~m}\) (B) \(40 \mathrm{~m}\) (C) \(20 \mathrm{~m}\) (D) \(80 \mathrm{~m}\)

Short Answer

Expert verified
The stopping distance when the automobile is travelling at 120 km/h will be 80 m

Step by step solution

01

Understand the problem

The vehicle is travelling at a speed of 60 km/h and stops within a distance of 20 m. If the speed is doubled, we need to determine the stopping distance.
02

Convert speed to a standard unit

Let's convert the speed from km/h to m/s. So, \(60 \mathrm{~km/h} = \frac{60 \times 1000}{3600} \mathrm{~m/s} = 16.67 \mathrm{~m/s}\). Likewise, \(120 \mathrm{~km/h} = \frac{120 \times 1000}{3600} =33.34 \mathrm{~m/s}\) .
03

Apply the physics of motion

The equation for the distance covered with constant deceleration is \( s= \frac{v^2}{2a} \) (where \(v\) is speed and \(a\) is constant deceleration). For a given deceleration, \(s\) is proportional to \(v^2\), this implies that for the same deceleration, if speed is doubled, distance \(s\) will increase by a factor of 4. Knowing that the original stopping distance was 20m at 60km/h, when the speed doubles to 120km/h, the stopping distance would quadruple.
04

Calculate the stopping distance

So, the new stopping distance will be \(20m \times 4 = 80m\).

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Most popular questions from this chapter

A car is moving on a circular path of radius \(100 \mathrm{~m}\). Its speed \(v\) is changing with time as \(v=2 t^{2}\), where \(v\) in \(\mathrm{ms}^{-1}\) and \(t\) in second. The acceleration of car at \(t=5 \mathrm{~s}\) is approximately (A) \(20 \mathrm{~ms}^{-1}\) (B) \(25 \mathrm{~ms}^{-1}\) (C) \(30 \mathrm{~ms}^{-1}\) (D) \(32 \mathrm{~ms}^{-1}\)

A glass wind screen whose inclination with the vertical can be changed is mounted on a car. The car moves horizontally with a speed of \(2 \mathrm{~m} / \mathrm{s}\). At what angle \(\alpha\) with the vertical should the wind screen be placed so that rain drops falling vertically downwards with velocity \(6 \mathrm{~m} / \mathrm{s}\) strike the wind screen perpendicularly? (A) \(\tan ^{-1}\left(\frac{1}{3}\right)\) (B) \(\tan ^{-1}(3)\) (C) \(\cos ^{-1}(3)\) (D) \(\sin ^{-1}\left(\frac{1}{3}\right)\)

A passenger reaches the platform and finds that the second last boggy of the train is passing him. The second last boggy takes \(3 \mathrm{~s}\) to pass the passenger, and the last boggy takes \(2 \mathrm{~s}\) to pass him. If passenger is late by \(t\) second from the departure of the train, then the find the value of \(2 \mathrm{t}\). (Assume that the train accelerates at constant rate and all the boggies are of equal length.)

A horizontal wind is blowing with a velocity \(v\) towards north-east. A man starts running towards north with acceleration \(a\). The time, after which man will feel the wind blowing towards east, is (A) \(\frac{v}{a}\) (B) \(\frac{\sqrt{2} v}{a}\) (C) \(\frac{v}{\sqrt{2} a}\) (D) \(\frac{2 v}{a}\)

A man wants to reach point \(C\) on the opposite bank of river. He can swim in still water with speed \(5 \mathrm{~m} / \mathrm{s}\) and can walk on ground with \(10 \mathrm{~m} / \mathrm{s}\). Velocity of river is \(3 \mathrm{~m} / \mathrm{s}\) and given \(A B=B C=500 \mathrm{~m} .\) $$ \begin{aligned} &\text { Column-I } & \text { Column-II } \\ &\hline \text { (A) The time to reach point } C \text { if } & \text { (1) } 125 \mathrm{~s} \\ &\text { he crosses the river by shortest } \\ &\text { path is } \\ &\text { (B) The time to reach point } C \text { if } & \text { (2) } 175 \mathrm{~s} \\ &\text { he crosses the river in shortest } \\ &\text { time. } \\ &\text { (C) The time to reach point } C \text { if } & \text { (3) } 120 \mathrm{~s} \\ &\text { he crosses the river along } & \text { (4) } 106 \mathrm{~s} \\ &\text { path } A C \text { (approximate value). } & \\ &\text { (D) Time to reach point } B \\ &\text { with zero drift } \end{aligned} $$
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