How Long Will it Take Me to Catch The Bus?

[bleedblue1234]

Homework Statement

Lisa is late for the school bus! (Oh - no) As Otto is pulling away with a constant acceleration of 1.10 m/s/s. Lisa is 11m away and running at a constant speed of 5.05 m/s..

When will Lisa catch the bus?

Where will Lisa catch the bus using the bus's initial position as the referance point?

Homework Equations

DeltaX= Vot+1/2a(t)sqrd

The Attempt at a Solution

Honestly I have sat here thinking for about 30 minutes... any insight on how to start would be great... (I don't just want the answer)

 

[songoku]

Hi bleedblue1234

I think the initial speed of the bus is zero. If I interpret the question correctly, it will be like this :

Assume that Lisa catches the bus at point X after both of them move for t seconds. At that time, the bus will travel distance d and Lisa will travel distance 11 + d.
Now set up two equations and solve for t and use the value of t to find where Lisa will catch the bus.

 

[tomcue]

I would approach this problem by first defining the variables and using the given information to set up a mathematical equation. In this case, the variables would be time (t), distance (x), initial velocity (Vo), and acceleration (a).

Using the formula DeltaX= Vot+1/2a(t)sqrd, we can plug in the given information to solve for the time it will take Lisa to catch the bus.

First, let's define the initial position of the bus as the reference point, meaning that x=0 at t=0. In this case, Lisa's initial position (xo) would be 11m away from the bus. So our equation becomes:

0=5.05t+1/2(1.10)(t^2)+11

Next, we can simplify this equation by multiplying 1/2(1.10)(t^2) to get:

0=0.55t^2+5.05t+11

Now we can use the quadratic formula to solve for t, which will give us the time it takes for Lisa to catch the bus:

t= (-5.05±√(5.05^2-4(0.55)(11)))/2(0.55)

Solving this equation, we get two possible times: t=1.50s or t=-4.55s. Since time cannot be negative, we can discard the negative solution and conclude that it will take Lisa approximately 1.50 seconds to catch the bus.

To determine where Lisa will catch the bus, we can plug this time back into our original equation to solve for the distance (x):

x=5.05(1.50)+1/2(1.10)(1.50^2)+11

Solving this equation, we get x=19.02m. This means that Lisa will catch the bus 19.02m away from its initial position.

I would also consider other factors that may affect this scenario, such as the precision of the given values and whether the bus is moving in a straight line or on a curved path. I would also suggest that Lisa try to run faster to catch the bus sooner!