IIT JEE , AIEEE Forum - Physics , Mechanics - a car travelling at 60km/h overtakes another car travelling at 42km/h.Assuming each car to

p.s.pandey

a car travelling at 60km/h overtakes another car travelling at 42km/h.Assuming each car to be 5m long,find the total road distance used for overtake

bhavanawannastudy

the question seems incomplete. how much is distance between the two cars before the overtake begins?

spideyunlimited

no it is not incomplete, it says find total distance used for overtake which means the distance covered on the road when the first car travels from the rear of the second car to its front that is 5m long, while the other car is ,of course, moving

digs2digi

is d answer 8.5 metre?????????

spideyunlimited

T = D / 42

T = D + 0.010 / 60

D / 42 = D + 0.010 / 60

60 D = 42 D + 0.420

18 D = 0.420

D =0.070/3 = .0233 km

= 23.33 m

add 10 m as this is the extra distance covered by overtaking car.

33.33 m

spideyunlimited

ALTERNATIVE METHOD

first using relative speed concept

T = D/S
T = 0.01 km (as 5 m of 1st car + 5m of 2nd car) / 60 - 42
T = 0.01 / 18

now this is the time for overtaking.
now to find total distance covered in this time
S = D / T
D = S x T
D = 60 km/h x 0.01 / 18
D = 0.0333 km = 33.33 m

dream

the relative speed of 2 car w.r.t. 1 is (60-48) km/hr = 18 km/hr = 5m/s

it has to cover a distance of 5 m
hence time required is 1 s
now it also has to get through its length of 5 m
hence it would take 1s more

so total time taken is 2 s
in 2 s 2nd car cover a distance of 60*5*2/18= 10*10/3 = 100/3 = 33 .33 m

hence my ans is 33.33 metre

bhavanawannastudy

the time required for the overtake will be 2 sec. and the rear end of the car moving with speed 60 km/hr will move 33.3 metre so my answer is 33.5 metre

elessar_iitkgp

Let the time of overtake be T. The overtake starts at the moment when the second car's front is just behind the first car's rear. The overtake ends when the first car's rear is in front of the second car's front. The distances moved by the cars are 60T and 42T
It can be seen by drawing a diagram that
60T - 42T = 10/1000
T = 1/1800 hrs = (1/1800)(3600) s = 2 s

The total road distance used for the overtake is = 60T + 5/1000 km = 60(1/1800) + 5/1000 km = 23/600 km = (23/600)(1000)

38 m.
You are all committing the same mistake ... the total road distance is asked and not the distance moved by the second car ... think carefully and try to understand the difference between them. If I remember correctly, this is a problem from HCV exercise ...

The road distance is the distance moved by the second car + the length of the second car itself. The question asks for the length of the road utilized and not the distance that the second car moves to overtake the first one.

iitkgp_bipin

relative velocity = 60-48 = 18 km/hr = 5 m/s

relative distance = 5+5 = 10 m

so time taken to overtake = 10/5 = 2 sec

distance traveled by 1st car in 2 sec = (60km/hr)(2 sec) = (50/3 m/s)(2 sec) = 33.33 m

rooney

@bipin sir

Let car travelling at speed 60 km/hr be B and that travelling at 42Km/hr be A.

So,33.33 m is the distance travelled by the behind most point of B between the time period when the front of B is touching with back of A and when back of B is touching the front of A. So, distance between the locations of the back of car B is 33.33 . However, we must add 5m to this since the length of car B is also using up the road space at the final position. So, we need the distance between the positions of back of B when it is just behind and front of B when it is ahead. So, answer should be 38.33m. As overtaking means completely being in front.

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