kinematics question

The car overtakes the other when the end of the 2nd car passes the front of the 1st car.
Let them be point A and B.
Distance between A and B =10m
Let t be the time taken to overtake the car.
Therefore,
60t=42t+0.01
=> t=1/1800hr or 2s
Therefore time required to overtake the car is 2 seconds.
Distance travelled in those 2s= 5 + 60*1/1800*1000
=5+60/1.8
=5+33.33
=38.33 m

time required fr overtaking is= L1+L2/V1-V2 =5+5 / 50/3-35/3 =10/5= 2sec
and total road distance =50/3 *2 = 100/3 m +5 m = 38.33m

relative velocity = 60 - 42 = 18 km/hr = 5 m/s
relative distance = 10m
therefore time taken = 10/5 = 2s

total road distance = (60*5/18) * 2 + 5m = 33.33m + 5m =38.33m

total road distance travelled by the car which is overtaked should be 33.5 y r u ading 5 metres??
nd total road distance travelled by the car which overtaked should be 43.5..

rel. velocity b/w the two is 5.

distance to be travelled by the first car is 10 m

t = 10 / 5 = 2
to cross by the second car

in this time first car moved 33.2 m

it also has to cover 5 m
so total 33.2 + 5 ~ 38m

I didn't understand why 5 mts. is getting added in calculation of to dist. of overtake ???

It is already included in the dist. covered in 2 secs ...

[The faster car had to overtake the length of the slower car and cover it's own length at the same time ... So 5 mts. is already included. So total dist. must be 33.33 mts.]

nope

it has to be included!!!

i dont know if it is right

v1=50/3
v2=35/3
t=x/(35/3)=(x+10)/(50/3)
x=23.33m
t=2 sec