I'm in a pickle with this question. Turning to anyone who can help me please.

A pedestrian is running at his maximum speed of 6.0 m/s trying to catch a bus that is stopped at a traffic light. When he is 16 m. from the bus, the light changes and the bus pulls away from the pedestrian with an acceleration of 1.0 m/s/s.

a) Does the pedestrian catch the bus and, if so, how far does he have to run? (If not, what is the pedestrian's distance of closest approach?)
b) How fast is the bus moving when the pedestrian catches it? ( or at the distance of closest approach)

I think I know that for the pedestrian to catch the bus, the displacements have to be the same, along with time. The pedestrian also has to make up for the 16m. so his displacement would be d+16....but I'm stuck, I don't know how to answer the question. Please Help. Thanks

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Integral
Staff Emeritus
Gold Member
It looks like you have the right idea for the distance. Now you just need to express the distance in terms of the knowns in your problem. You should have seen the equations you need in class or in your text.

What have you been given to work with?

Haha, the problem is that I have a similar question sitting in my locker at school, and this question is due tomorrow, I'm just so confused, because I seems like the answer is right there. I have many equations to work with, but I seem to have 2 variables for both the bus and the pedestrian. If you could suggest a way to either get the time or the displacement, I could work off of that.

pedestrian : velocity = 6.0m/s
Acceleration = 0 m/s/s (constant velocity, not sure if correct)
displacement = displacement of bus + 16m
time = same as bus

bus: acceleration = 1.9 m/s/s
V(initial) = 0m/s
V(final) = ?
displacement = ?
time = ?

The velocity (final) for the bus would have to be equal to the pedestrian (thinking out loud, stop me if I'm wrong). therefore V(f) = 6.0 m/s

You then use the equation v(f)^2 = v(i)^2 + 2ad
(6.0m/s)^2 = 2(1.0m/s/s) (d)
36 m/s = 2.0 m/s/s (d)
18 m = displacement of bus

therefore 18 + 16 = displacement of person
34 m = displacement of person.

thats definately wrong

Päällikkö
Homework Helper
You did the bus part right, but the displacement of the pedestrian should have a -16, as he starts behind the bus. Why should the displacement of the person have anything to do with the velocity of the bus (your 18 + 16) ?

The problem gets really simple if you think about curves.

Here's what you do. Write down your givens first.

My initial velocity vi(me) = my final velocity = 6.0m/s
acceleration of the bus a(bus) = 1.0m/s/s
my initial position di(me) = 0
the bus's initial position di(bus) = 16m
my final displacement df(me) = ?
the bus's final displacement df(bus) = ?
df(me) = df(bus) = df
my time = the bus's time t(me) = t(bus) = t

(sorry don't really know how to make abbreviations on the computer)

df(bus) = di(bus) + Vi(bus)t + 1/2(a(bus))t^2
since vi(bus) = 0
df = di(bus) + 1/2(a(bus))t^2

df(me) = vi(me)t *because the person running is not accelerating

since df(bus) = df(me)

di(bus) + 1/2(a(bus))t^2 = vi(me)t

i won't do it, but you know all the variables except time. good luck!

The problem will have 2 solutions. if the passenger chasing the bus makes it then it is a solution. And when he continue to run past the bus, the bus will overtake him (provided the bus continues to accelerate and that the bus speed is less than 6m/sec at the first meeting)

So the solution must involve some quadratic equation.

Yes, the passenger can catch the bus after running 24 meters
The bus was traveling at 4m/sec (below the speed of the passenger)
It took 24/6 = 4 secs to catch the bus.

Another solution: (we are not much interested in this solution because the passenger boards the bus when he meets the bus and does not continue to run)

If he continue to run at 6m/sec, the bus would accelerate past him from 4 m/sec and 4 secs later than the first meeting, the bus would pass him.

Delphi51
Homework Helper
Oh, great, we're all enjoying a very nice high school problem!
It involves uniform motion (pedestrian). Formula: d = vt
It involves constant accelerated motion (bus). Formulas v = at,
d = ½at².
One of the d formulas should have 16 added to it to make the two d's measure from the same position. The two d formulas give the positions of the pedestrian and bus at any time in the future, taking t=0 to be when the bus takes off.

One approach from here would be to graph the two d functions of time.
Another would be algebra to solve the system of equations for the quantity you want.
Another would be a table of distances for each for every second of time.

hello dear. i dont know whether u are still stuck or not, but here is a solution to your problem

Consider that the person catches the bus after t seconds and while the time interval t, the bus has covered S meteres.

Now for the pedestrain.....
Constant Velocity=Average Velocity, V=6 m/s
Time taken = t sec
Distance or doisplacment covered = 16 + S meters

The equation S=Vt will give
S + 16 = 6 x t..............(1)

For the Bus

Initial Velocity, Vi = 0 m/s
Acceleration, a = 1 m/s/s
Time to cover S meters = t sec
displacment, S = S meters

Equation, S=(Vi x t) + (0.5x ax t^2)
S = 0.5 x 1 x t^2.. as Vi = 0 m/s
S = 0.5 t^2..............(2)

equation (1) states S = 6t - 16.....put in equation (2)
Result is
6t - 16 = 0.5t^2
solve the above quadratic equation , you will get t = 4 seconds and t= 8 sec....

Conclusion: After 4 seconds the pedestrain just reaches the bus, if he continue racing with the bus he overtakes it. but after 8 seconds the bus reaches the pedestrain and will overtake him.

Assumptions made: The pedestrain Velocity is constatn throughout and acceleration of the bus is also constant.

if you wish to calculate the distance covered by the bus or the pedestrain, just put time t = 4 sec in equation (1) and (2) you will get distance covered by the pedestrain in meters before catching the bus Last edited: