Answer to Question #129419 in Classical Mechanics for Touheed Iqbal

Question

A traffic warden starts from rest at Shamsabad Murree Road 3second after when a one wheeler crosses him speeding at a constant rate of 110 fps. The warden accelerates at the rate of 7m/s2 and reaches his speed of 100 km/h which he maintains for 5 seconds when suddenly his bike stops due to uncertain reasons. He then again accelerates with the same rate and reaches to the speed of 150km/h. Determine how long from Shamsabad he will catch the one wheeler.

Answer 1

As per the given question,

Initial speed of traffic warden (u)=0

At, t=3sec, speed of one wheeler (v)=110 fps= 33.5m/sec

Acceleration of traffic warden$(a)= 7 m/sec^2$

Time$(t_1)$ taken by the warden to reach the speed $(v)= 100 km/hour= \frac{450}{9} m/sec$

$v=u+at_1$

$\Rightarrow \frac{450}{9}=0+7t_1$

$\Rightarrow t_1=\frac{450}{63} sec$

time spend at the time of break = 5sec

again, time taken by the object to approach the velocity 150 km/hour$=\frac{125}{3} m/sec$

$t_2=\frac{125}{3\times 7 }=\frac{125}{21}sec$

total distance covered by the traffic warden, from the contact to one wheeler,

$v^2=u^2+2as_1$

$\Rightarrow s_1=\frac{v^2-u^2}{a}=\frac{(\frac{450}{9})^2}{7}= 357.14 m$

distance covered by the traffic warden till 3 sec,

$\Rightarrow s=\frac{7\times 3\times 3}{2}= 31.5m$

$\Rightarrow d_1= (357.14- 31.5)m =325.64m$

distance covered by the traffic warden in second starts,

$d_2=\frac{7\times (\frac{125}{21})^2}{2} =124m$

Hence, total distance covered by the traffic warden after crossing of one wheeler $=d_1+d_2 =(124+325.64)m =449.64m$

total distance covered by the one wheeler after crossing the traffic warden$=v(t_1+5+t_2 -3)=33.5(\frac{450}{63}+2+\frac{125}{21}) m$

$=505.51m$

Hence, the required difference $=(505.51-449.64) m=55.87m$