# Catching a bus.

## Homework Statement

A man runs at a speed of 4.0 m/s to catch a standing bus. When his distance to the door is 6.0m, the bus starts moving forward and continues with a constant acceleration of 1.2 m/s2.

## Homework Equations

a) How many seconds does it take for the man to reach the door ?
b) If at the beginning he is 10.0 m from the door, will he (running at the same speed) ever catch up? Simple answer is not enough - explanatory answer using calculations or graphical method is needed

## The Attempt at a Solution

a)
xm= 4*t + 6
xb=1/2*1.2*t2

xm= xb
0.6*t2+-4t-6=0
delta=30.4
t1= -1.26s (rejected because it's <0)
t1= 7.92s (accepted because it's >0)

yeah, the method is fine....

i wasn't sure when i added 6m to the equation of man.

ehild
Homework Helper
You have to subtract. The positions of both the bus and man are the same when the man catches the bus. If you write the position with respect to the initial place of the bus, the man is initially behind by 6 m. So its position is xm=4t-6.

ehild

ehild's right
subtract it

let the total distance covered by man & bus (from their initial position) is x(m) & x(b)

so x(m) = x(b) + 6

Yes indeed, I think i get it now.

If we write both equations with respect to the man, they should be:
xbus = 1/2*a*t2+6
xman = 4t

If we write both equations with respect to the bus, they should be:
xbus = 1/2*a*t2
xman = 4t-6

And then, we will get the same equation which is: 0.6t2-4t+6=0
But after solving it, we get two values, t1 = 2.279s and t2 = 4.38s

How can we know which of these values is the right one :yuck:

ehild
Homework Helper
Funny, but both ones.

ehild

What do you mean

ehild
Homework Helper
Think. You can catch the bus when it just have started and and its speed is lover than yours. You can run further and leave it behind you, but it accelerates and will catch you a later instant of time. Choose the shorter time.

ehild

Is there any way to verify this mathematically or we just take the small value for this kind of situation!
Why we reject the value of 4.38s ?

Last edited:
ehild
Homework Helper
v(bus)= at, so the bus reaches the speed of the man after 4/1.2 = 3.33 s. The shorter time is needed for the man to catch the bus, but that time the bus travels slower than the man. If the man runs further he leaves the bus behind himself. But the bus gains speed further, and after 4.38 s, it will catch the man. But the question was: "How many seconds does it take for the man to reach the door ?"

ehild

ok, well, this imagination wouldn't come to my mind
but here how i can see it:
we substitute 2.279s and 4.38s in Vbus to see if it's >4m/s or if it's <4m/s.
so for 2.279s ......Vbus=2.7348 m/s and that's slower than the speed of man, so he will catch the bus.
and for 4.38s Vbus=5.256 m/s and that's > speed of man...so he will not catch the bus at that time.

ehild
Homework Helper
It is not the speed that counts. It is the distance they covered.
In 4.38 s the bus travelled 11.5 m, and the man did 17.5 m, 6 m longer than the bus. They are at the same place, the man can jump at the back door (if it is open).

In 2.28 s time the bus travelled 3.12 m distance and the man did 6 m longer, 9.12 m. They are at the same place, the man can jump in the bus.

See what happens if the man is 10 m behind the bus when it starts.

ehild

In case he was 10m away, the equations would be:
xm=4*t
xb=0,6*t2+10

and then, we will have 0.6t2-4t+10=0
Delta = -8 and that's <0
so there is no solution for this equation, meaning that the poor man will never catch the bus

ehild
Homework Helper
That is life... twice or never :) And what happens if the man is 20/3 meters behind the bus?

ehild

in that case, we will have 0.6t2-4t+20/3 = 0
Delta= 0
so he will catch the bus at t=4/1.2=3.33s

ehild
Homework Helper
in that case, we will have 0.6t2-4t+20/3 = 0
Delta= 0
so he will catch the bus at t=4/1.2=3.33s

And he has one chance only. Next time do this calculations before a starting to run to catch a bus

ehild

haha