A car travelling at 60 Km/h overtake another car travelling at 42Km/h .Assuming each car to be 5.0m long,find the tame taken during the overtake and the total road distance used for the overtake
30The 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
1time 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
1relative 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
1total 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..
6rel. 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
49bhai 1 suggestion!!
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Title: kinematics.
Message: A car travelling at 60 Km/h overtake another car travelling at 42Km/h .Assuming each car to be 5.0m long,find the tame taken during the overtake and the total road distance used for the overtake??
1I 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.]
1i 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
