Calculating Passing Time of Accelerating Car and Bus

In summary, a car and a bus are traveling in the same direction, with a distance of 20m between them. The car has a velocity of 10m/s and starts accelerating at 5m/s^2, reaching a maximum velocity of 35m/s before stopping its acceleration after 5 seconds. The bus has a current velocity of 20m/s. After 5 seconds, the car has traveled a distance of 112.5m and the bus a distance of 100m. The car and bus are now 7.5m apart, with the car traveling at a constant speed of 35m/s. It takes the car 6.6 seconds to completely pass the bus, with a relative speed
  • #1
1irishman
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0

Homework Statement


A car of length 4m and a bus of length 12m are traveling in the same direction 20m apart. The cars current velocity is 10m/s and the buses current velocity is 20m/s, the car suddenly starts to accelerate at 5m/s^2 and stops accelerating after 5 seconds. How long does it take the car to completely pass the bus?

Homework Equations


d=Vit + 1/2at^2
i'm not sure about other ones that can be used here...

The Attempt at a Solution


I figured (not sure if it's right though) the bus travels 100m in 5s and the car travels 112.5m in the same amount of time. So the original distance of 20m between the car and the bus has lessened to 20m - 12.5m = 7.5m. So the distance between the bus and the car is now 7.5m. Not sure how to proceed from here...hints? Thanks.
 
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  • #2
I see no mention of the 2 seconds in the question.
The 112.5 m is correct for the car at time 5 s. Good start.
 
  • #3
i figured (not sure if it's right though) the bus travels 100m in 5s and the car travels 112.5m in the same amount of time. So the original distance of 20m between the car and the bus has lessened to 20m - 12.5m = 7.5m. So the distance between the bus and the car is now 7.5m. Not sure how to proceed from here...hints? Thanks.
 
  • #4
Well, the car stopped accelerating after 5 s, so both vehicles must now be in motion at constant speed. The next bit should be easier . . .
Don't forget the lengths of the vehicles!
 
  • #5
after 5s the new constant speed of the car should be 35m/s right? And the total distance should now be 23.5m right?
 
  • #6
okay...i think I've solved it:
since cars new constant speed is 35m/s and distance is 23.5m.
The relative speed to the bus should be 15m/s.
So, v=15m/s
d=23.5m
then t=1.6s
So, it should take the car 6.6s to pass the bus right?
 
  • #7
Very nicely done! Congrats.
 
  • #8
Thanks!
 

1. What is relative velocity?

Relative velocity is the velocity of an object in relation to another object. It takes into account the motion of both objects and is measured as the change in position over time.

2. How is relative velocity different from absolute velocity?

Absolute velocity is the velocity of an object in relation to a fixed point, while relative velocity is the velocity of an object in relation to another moving object. Absolute velocity is constant, while relative velocity can change depending on the motion of the reference object.

3. How do you calculate relative velocity?

Relative velocity can be calculated by subtracting the velocity of one object from the velocity of another object. For example, if object A is moving at 10 m/s and object B is moving at 5 m/s, the relative velocity of object A with respect to object B is 5 m/s.

4. What is the importance of understanding relative velocity?

Understanding relative velocity is important in many scientific fields, such as physics, astronomy, and engineering. It helps us understand the motion of objects in relation to each other and is crucial in predicting and analyzing the behavior of systems.

5. What are some real-world applications of relative velocity?

Relative velocity is used in many real-world applications, such as calculating the speed of airplanes or ships in relation to the ground, predicting the trajectory of a projectile, and understanding the motion of celestial bodies in space.

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