I don't necessarily want the answer (at best ill use it to check my own answer). I'm trying to figure out how to work this out. so far ive been working only questions with only one moving object, this is obviously different and im completely stumped

Your are sitting in your car, which is standing at the side of the road with the engine idling. In the rearview mirror, you see one of your class mates approaching on a bicycle on the sidewalk. The moment the other student passes you, you start accelerating at 3.36 m/s2, while the bicycle continues to move along at a constant speed of 10 m/s. After what distance will you catch up with your classmate?

Using kinematic equations: Surely your teacher has given you 4 equations with position, time, velocity, and acceleration?

x = vi*t +.5*a*t^2

Your position = 0*t +.5*3.36*t^2

His position = 10*t +.5*0*t^2


Set the two position equations equal to each other and solve for time. Plug time into the original position equation

d_bean = 1/2a_bean * t^2 + v_bean*t + x_bean
d_fred = v_fred * t

bean distance = fred distance

1/2a_bean * t^2 + v_bean*t + x_bean = v_fred * t

v_bean = 0
x_bean = 0

so:
1/2 * a_bean * t^2 = v_fred * t

solve for t so you know how long it took to travel, then plug it back into either equation for distance

how come you remove the units from 3.36m/s^2?

he gave us 3 equations though i think?

You wouldn't remove any units, I just didn't type them out. Everything should work unit wise with the values they gave you.

im messing something up.. fractions still screw me up, even after calculus and trig >_<

1/2(3.36m/s^2)t = 3.36mt/2s^2?

Yes, except that the equation is t^2 not just t

(.5) * (3.36 m/s^2) * t^2

= (1.18 m/s^2) * t^2

So after you solve for t and plug it back into the equation the s^2 will cancel out and you're left with just distance

so i have

car: 1.68mt^2/s^2
bike: 10mt/s

what exactly did i just find? also what am i doing after that?

edit: after making them equal i have t = 5.9524s

This post was edited by Bean` on Jan 25 2015 04:46pm

Those are the equations for their positions. Where t = Time

You are looking for the point where they are at the same distance, so set the two equations equal to each other. Solve for t and you will have how long it takes for the car to catch up to the bike.

Now that you know what time is, you can plug it into one of those two equations and solve for distance. This will be the distance at which the car has caught the bike

This post was edited by RzChaos on Jan 25 2015 04:47pm

yeah i got t = 5.9524s

but what equation does that go into? x = x not + v not t + 1/2 at^2?

if so then what do i put for the variables since i have two different moving objects?