PLZ PTOVIDE THE SOLUTION FOR QUES NO. 21,22 PAGE NO. 52 AND QUE NO. 28,31 PAGE NO 53 OF HCV PART 1   KINEMATICS.......

RATES ASSURED FOR CPRRECT ANS AND EXPLANATION

Could u post the question please ?

21) V_P = 25 m/s.

V_C = 20 m/s.

In 10 sec culprit reaches at point B from A.

Distance converted by culprit S = vt = 20 × 10 = 200 m.

At time t = 10 sec the police jeep is 200 m behind the

culprit.

Time = s/v = 200 / 5 = 40 s. (Relative velocity).

In 40 s the police jeep will move from A to a distance S, where

S = vt = 25 × 40 = 1000 m = 1.0 km away.

22) v₁ = 16.6 m/s.

v₂ =  11.6 m/s.

Relative velocity between the cars = (16.6 – 11.6) = 5 m/s.

Distance to be travelled by first car is 5 + 5 = 10 m.

Time = t = s/v = 10/5 = 2 sec to cross the 2nd car.

In 2 sec the 1st car moved = 16.6 × 2 = 33.2 m

H also covered its own length 5 m.

Therefore total road distance used for the overtake = 33.2 + 5 = 38 m.

28) s = ut + 0.5*at²  S = 12 m  a = 9.8 m/s²  u = 0  implies  t = 1.5 sec . For cadet velocity = 6 km/hr = 1.6 m/sec     S = vt = 1.57 × 1.66 = 2.6 m.

31 ) As u = 0  &  elevator descends downward  implies coin has to move more distance than 1.8 m to strike the floor  & T = 1 sec.  S_c = ut + 0.5*at² = 0 + 1/2 g(1)² = 0.5 g   &   S_e = ut + 0.5*at²   =   u + 0.5 a(1)² = 0.5 a  Total distance covered  1.8 + 0.5 a = 0.5 g  therefore  1.8 + 0.5a = 0.5 g implies  a = 6.2 m/s² = 6.2 × 3.28 = 20.34 ft/s².