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Chapter 2: Problem 17

A bus is moving with a velocity \(10 \mathrm{~ms}^{-1}\) on a straight road. A motorist wishes to overtake the bus in \(100 \mathrm{~s}\). If the bus is at a distance of \(1 \mathrm{~km}\) from the motorist, with what velocity should the motorist chase the bus? (A) \(50 \mathrm{~ms}^{-1}\) (B) \(40 \mathrm{~ms}^{-1}\) (C) \(30 \mathrm{~ms}^{-1}\) (D) \(20 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The required velocity of the motorist to overtake the bus in 100 seconds is \(20 ms^{-1}\) (Option D).

Step by step solution

01

Determine the total distance motorist must travel

To overtake the bus, the motorist must cover an extra distance traveled by the bus in the given time. Let's first find the distance traveled by the bus in 100 seconds: distance = velocity 脳 time, so: distance_BUS = 10 m/s 脳 100 s = 1000 m Now, let's add the distance between the motorist and the bus (1km = 1000m) to the distance traveled by the bus: total_distance = 1000m + 1000m = 2000m
02

Calculate the required velocity

Now, we can find the motorist's required velocity by dividing the total distance by the time it took to overtake the bus: velocity = total distance / time, so: velocity_motorist = total_distance / time = 2000m / 100s = \(20 ms^{-1}\)
03

Choose the correct answer

Comparing the calculated velocity of the motorist with the given options, we can see that the correct answer is: (D) \(20 ms^{-1}\)

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Most popular questions from this chapter

A particle is moving on a circular path of radius \(\frac{100}{\sqrt{19}} \mathrm{~m}\) in such a way that magnitude of its velocity varies with time as \(v=2 t^{2}+t\), where \(v\) is velocity in \(\mathrm{m} / \mathrm{s}\) and \(t\) is time in \(s .\) The acceleration of the particle at \(t=2 \mathrm{~s}\) is (A) \(21 \mathrm{~m} / \mathrm{s}^{2}\) (B) \(9 \mathrm{~m} / \mathrm{s}^{2}\) (C) \(10 \mathrm{~m} / \mathrm{s}^{2}\) (D) \(13.5 \mathrm{~m} / \mathrm{s}^{2}\)

A road is \(5 \mathrm{~m}\) wide. Its radius of curvature is \(20 \sqrt{6} \mathrm{~m}\). The outer edge is above the inner edge by a distance of \(1 \mathrm{~m}\). This road is most suited for a speed \(\left(g=10 \mathrm{~ms}^{-2}\right.\) ) (A) \(10 \mathrm{~ms}^{-1}\) (B) \(10 \sqrt{5} \mathrm{~ms}^{-1}\) (C) \(100 \mathrm{~ms}^{-1}\) (D) \(40 \sqrt{6} \mathrm{~ms}^{-1}\)

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If \(\vec{r}=b t \hat{i}+c t^{2} \hat{j}\) where \(b\) and \(c\) are positive constants, the velocity vector make an angle of \(45^{\circ}\) with the \(x\) and \(y\) axes at \(t\) equal to (A) \(\frac{b}{2 c}\) (B) \(\frac{b}{c}\) (C) \(\frac{c}{2 b}\) (D) \(\frac{c}{b}\)

For a particle in uniform circular motion, the acceleration \(\vec{a}\) at a point \(\mathrm{P}(R, \theta)\) on the circle of radius \(R\) is (Here \(\theta\) is measured from the \(x\)-axis) (A) \(-\frac{v^{2}}{R} \cos \theta \hat{i}+\frac{v^{2}}{R} \sin \theta \hat{j}\) (B) \(-\frac{v^{2}}{R} \sin \theta \hat{i}+\frac{v^{2}}{R} \cos \theta \hat{j}\) (C) \(-\frac{v^{2}}{R} \cos \theta \hat{i}-\frac{v^{2}}{R} \sin \theta \hat{j}\) (D) \(\frac{v^{2}}{R} \hat{i}+\frac{v^{2}}{R} \hat{j}\)
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