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Two objects two differents accelerations

  • Thread starter 67NICK67
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  • #1
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I know this is a repost but i need this for tomorow morning so id greatly appriciate it if someone could help me out, thanks a bunch!!!

Homework Statement



A personne walking at a speed of 1.7m/s see's a stoped bus 25m from him. At this moment this person starts to run with an acceleration of 1,3m/s2. After 4 secondes this personne has reached his maximum speed et it stays constant, but at the same time the bus starts moving and accelerates 2,6m/s2.

Does this person suceed in catching the bus? If so, at what time and at what distance, if not what was the smallest distance between the person and the bus.

- Im sorry if there any mistakes, i translated from french.


Homework Equations



Thats my problem i dont know what equations takes into account two moving objects


The Attempt at a Solution



Ive got this so far:

Alright well, by using the initiale speed of 1,7m/s and 4s, i found his constante speed of 6.9m/s. So after that i used the acceleration and his speed to determine the distance he travelled, 17,2m. i subtracted 17,2m from 25m, which left 7,8m between the bus and the person. Then im stuck, how do i incorporate both speeds in one equations to see if he caught the bus or not?

Update, i tryed using a time of 8s for both the peron and the bus, and i see that the bus is goin faster then the person and has travelled a greater distance, the only problem is dont know what has happened between 4s and 8s, im completly lost
 

Answers and Replies

  • #2
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You first need to find the distance he traveled before his speed was constant.

Once you find that distance then find the relative distance between him and the bus as a function of time. He travels at some constant speed v*t towards the bus which closes the gap while the bus moves away at the speed of 1/2 a*t^2.

Ignoring the movement of the bus for a moment the distance between the runner and the bus is decreasing at the rate of v*t.

Ignoring the movement of the person for a moment the bus is moving at a distance of 1/2 a*t^2 away from the person, thus increasing the distance.

Combining these we have the total distance is the distance between the bus and the person at time 0. Plus the distance the person is traveling. Minus the distance the bus is traveling.

What we end up with is a quadratic function that looks like: D = d - v*t + 1/2 a*t^2. Where 'd' is the distance between the bus and the person at time t = 0 (actually the fourth second of the problem but the zeroth of the equation) and v is the velocity of the person, and a is the acceleration of the bus.
 
Last edited:
  • #3
5,428
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This is a tough problem and you will need to solve equations.

One approach is to assume that the runner catches the bus at x metres from the point it was stopped, in time t after it started. Now you can write an equation in x and t for runner and for the bus. If the runner can catch the bus, you will be able to solve for x and t from the two equations. If the runner can't catch the bus, the equations will have no finite solutions.

For the bus the equation is

x = 1/2 (2.6) t^2

try and write the same thing for the runner.
 
  • #4
5
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Thank you sooooo much, that simple little question has been driving me insane, thanks again!!!!

nick
 

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