Q. The driver of a car moving with a speed of $10ms^{- 1}$ sees a red light ahead, applies breaks and stops after covering 10m distance. If the same car were moving with a speed of $20\ ms^{- 2}$, the same driver would have stopped the car after covering 30 m distance. Within what distance the car can be stopped if travelling with a velocity of 15$ms^{- 1}?\ \ $Assume the same reaction time and the same deceleration in each case.
Explanation
(a) If $t_{0}$is the reaction time, then the distance covered during decelerated motion is $10 - 10t_{0}$.
Now, in the first case,
$10^{2} = 2a(10 - 10t_{0})$ ...(i)
Similarly, in the second case,
$20^{2} = 2a(30 - 20t_{0})$ ...(ii)
Again, in the third case,
$15^{2} = 2a(x - 5t_{0})$ ...(iii)
Dividing Eq.(ii) by Eq. (i),$\ \frac{20^{2}}{10^{2}} = \frac{30 - 20t_{0}}{10 - 10t_{0}}$
or $40 - 40t_{0} = 30 - 20t_{0}$
or $20t_{0} = 10$ or $t_{0} = \frac{1}{2}s$
Dividing Eq. (iii) by Eq. (i), we get
$\frac{225}{100} = \frac{x - 15t_{0}}{10 - 10t_{0}}$ or $\frac{9}{4} = \frac{x - 15 \times \frac{1}{2}}{10 - 10 \times \frac{1}{2}}$
$45 = 4x - 30$ or $4x = 75$
or $x = \frac{75}{4}m = 18.75$m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.