Calculating Time and Distance for Car Overtaking at Different Speeds

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A car traveling at 60 km/h overtakes another car at 42 km/h, with each car measuring 5 meters in length. The relative speed during the overtaking maneuver is calculated to be 18 km/h (or 5 m/s), allowing the overtaking car to cover a total distance of 10 meters in 2 seconds. However, the total road distance required for the overtaking maneuver should account for the length of the overtaken car as well, leading to a calculated distance of 34 meters when factoring in the speed and time. The discussion highlights the importance of considering absolute speed and additional lengths in determining the total distance for safe overtaking. Clarification on safety distance is deemed unnecessary since it was not specified in the problem statement.
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Homework Statement



A car traveling at 60 kmph overtakes another car traveling at 42 kmph. Assuming each car to be 5.0 m long, find the time taken during the overtake and the total road distance used for the overtake.[/B]

Homework Equations


Relative Velocity of a body A with respect to another body B = velocity of A - velocity of B

The Attempt at a Solution



Velocity of Faster Car with respect to slower car = 60 - 42 = 18 kmph = 5mps.

Total distance to cover = 5 m + 5 m = 10 m

Total time taken for overtake = 10/5 = 2 s.

But I do not understand the second part of the question. Kindly help.[/B]
 
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What is the distance between [the back of the overtaking car before] and [the front of the overtaking car afterwards]?

(some safety distance should be kept for real overtaking maneuvers)
 
mfb said:
What is the distance between [the back of the overtaking car before] and [the front of the overtaking car afterwards]?

(some safety distance should be kept for real overtaking maneuvers)
okk.. but there is nothing in the question about the safety distance ? is that information necessary to solve this question ? the distance between [the back of the overtaking car before ] and [the front of the overtaking car afterwards] is 15 m.
 
If the problem statement does not say anything about safety distance and if you also did not consider it in previous problems, ignore it.

anandtr2006 said:
the distance between [the back of the overtaking car before ] and [the front of the overtaking car afterwards] is 15 m.
The car moves at much more than 10 meters per second. How can this be true, given that the car moves for two seconds?
 
mfb said:
If the problem statement does not say anything about safety distance and if you also did not consider it in previous problems, ignore it.

The car moves at much more than 10 meters per second. How can this be true, given that the car moves for two seconds?
I don't quite get your question. the faster car's real speed is 16.7 m/s. But since relative velocity is taken, the velocity of the car relative to the slower car is 5 m/s. Thus it can travel 10 metres in only two second ? Am I right ?
 
The second part of the question asks about the distance on the road, and the car travels at 60km/h relative to that road. How long does the road have to be? The absolute speed matters here.
 
mfb said:
The second part of the question asks about the distance on the road, and the car travels at 60km/h relative to that road. How long does the road have to be? The absolute speed matters here.
so overtaking distance is the length of the road which needs to be traveled by the faster car in order to overtake the slower car.
distance on the road = velocity with respect to road x time of overtaking
d = (17 m/s) x 2 = 34 m.
so 34m should be the answer. right ?
 
anandtr2006 said:
so 34m should be the answer. right ?
Well, this would be commonly called the distance the car travelled, but you are asked for "the total road distance used for the overtake" which probably includes one more car length.
 
In addition, be careful with rounding errors.
 
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