Question

The drive of an express train suddenly sees the red light signal 50m ahead and applies the brakes. If the average deceleration during the braking is $10.0m{{s}^{-2}}$and the reaction time of the driver is 0.75 sec, the minimum speed at which the train should be moving so as not to cross the red signal is
A. 27km/hr
B. 144km/hr
C. 72 km/hr
D. 83 km/hr

Answer 1

Hint: The question exercises the concepts of Newton's laws of motion. There are two specific newton’s laws of motion, which are required to solve this problem. We will be needing these two equations of rectilinear motion: $s=ut+\dfrac{1}{2}a{{t}^{2}}$and ${{v}^{2}}-{{u}^{2}}=2as$, where all the alphabets mean their general variable values.

Step by step solution:
Let’s start by understanding what is given in the question. We have a red light signal at a distance of 50m from the current position of the train. Hence, the stopping distance ‘${{S}_{d}}$’ is 50m. The average deceleration during braking is given, hence the braking acceleration (a) is, $a=-10.0m{{s}^{-2}}$. We are also given that the reaction time of the driver is 0.75 sec. This is the amount of time the driver takes to push the brakes since the time he saw the red light.

Due to the reaction time, the train moves a certain distance (${{S}_{1}}$) with its initial velocity (u). Since, no initial acceleration is mentioned; we will consider initial acceleration to be zero.

Therefore, using $s=ut+\dfrac{1}{2}a{{t}^{2}}$, we find the distance ${{S}_{1}}$to be, ${{S}_{1}}=u(0.75)+\dfrac{1}{2}(0){{(0.75)}^{2}}\Rightarrow {{S}_{1}}=u(0.75)m$.
Therefore, the updated braking distance becomes, ${{S}_{d}}=50-{{S}_{1}}=50-0.75(u)$.

We know that upon braking the final velocity of the train will be zero. Therefore (v=0). Further, the braking acceleration is, $a=-10.0m{{s}^{-2}}$. Substituting these values in the formula, ${{v}^{2}}-{{u}^{2}}=2as$, we get, ${{(0)}^{2}}-{{u}^{2}}=2(-10){{S}_{d}}\Rightarrow -{{u}^{2}}=-20(50-0.75u)\Rightarrow {{u}^{2}}+15u-1000=0\to (1)$

To solve this equation of second order, let’s consider the coefficients of the variable u. a=1, is the coefficient of ${{u}^{2}}$. B=15, is coefficient of $u$ and c=-1000, is the constant.

Using the formula to find the roots of a second order equation, $a{{x}^{2}}+bx+c=0\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Therefore, the value of ${{b}^{2}}-4ac$ is, ${{15}^{2}}-4(1)(-1000)=225+4000=4225$.

Hence, the value of the initial velocity can be, $u=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\Rightarrow u=\dfrac{-(15)\pm \sqrt{4225}}{2(1)}\Rightarrow u=\dfrac{-(15)\pm 65}{2}$. Hence, the two possible values of ‘u’ are 25 (or) -40.

For the initial velocity train being, 40m/s, the train’s speed will be, $40m{{s}^{-1}}=40\times \dfrac{3600}{1000}=144kmh{{r}^{-1}}$. Hence, option B is the solution.

Note:
One of the shortcut tricks to convert any speed, x m/s to km/h is by multiplying the value ‘x’ with $\dfrac{18}{5}$. Therefore x m/s is $\dfrac{x18}{5}$kmph.
Similarly, to convert any speed, x km/h to m/s, we must multiply the value ‘x’ with $\dfrac{5}{18}$. Therefore x kmph is $\dfrac{5x}{18}$m/s.