# Motorcycle Catches a Car

1. Sep 6, 2004

### EaGlE

A motorcycle is 27 m behind a car that is traveling at constant speed on a straightaway. Initially, the car and the motorcycle are both traveling at the same speed, $$Vcar$$ . At some time $$t1$$, the motorcycle starts to accelerate at a rate of $$a=6m/s^2$$ . By the time the motorcycle catches up with the car, at time $$t2$$ , it is traveling at twice the car's speed.

A.) How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find $$(t1-t2)$$. Express the time numerically in seconds.

B.) What is the initial speed of the car? Express the speed numerically in meters per second. answer must be in m/s

C.) How far did the motorcycle travel from the moment when it starts to accelerate (at time t1 ) until it catches up with the car (at time t2)? Answer numerically in meters.

ok what i really get stuck on is what formula should i use? i really dont know how to start problems like this, because it's im always like "What formula should i use?" can someone help me start this question and tell me your secret on how you know what formula you use on any physics problem? i like physics and i think it's awesome how it works, but it's so hard for me.

2. Sep 6, 2004

### dink

What you need is the basic idea of acceleration, velocity, and position with respect to time. With some integrations holding a as constant you end up with:

V(final) = V(intial) + AT (used for maximum velocity)
x(final) - x(initial) = V(intial)T + (1/2)A(T^2) (distance formula)

Using these as a system of equations you can solve a, b, and c by substituting them into eachother. Hoping this helps.

3. Sep 6, 2004

### Alkatran

The best way to solve a problem is to:
A) Write down all the units you have, distance, time, etc.
This saves you from having to interpret the question 50 times
*Know the relations between the units (aka formulas)
Speed is distance over time, but this is obvious because of their units (m/s, m, and s)

B) Figure out as many formulas as you can, for example:
By the time the motorcycle catches up with the car, at time $$t2$$ , it is traveling at twice the car's speed.
means that the final speed of the motorcycle is.. twice the car's speed!
V(final,motorcycle) = 2 * V(Car)

C) Expand the units you know.
This is where it gets tricky. Personnaly, I like to ignore the question and solve all the units in a question before looking at the question and putting a box around the answer. This means every question is the same to me, and repetition = .
Now, the trick to this is knowing the formulas and relations. If you know time and distance you can find average speed, if you know time, initial speed, and final speed, you can find acceleration. If you know time, acceleration and initial speed you can find final speed. ETC

So my suggestion to you is this: Write down the units you know AND the ones you don't:

Everything
t = ?

Motorcycle
d = ? (minimum 27)
v(i) = v(c)
v(f) = 2*v(c)
change of v = v(c)
a = 6 m/s^2

Car
d = ?
a = 0
v(i) = v(f) = v(c) = ?

Look at the motorcycle list. You know the total change of speed over time, and acceleration (the average change of speed over time). What can you find?

4. Sep 6, 2004

### EaGlE

ok lets work on problem A. first.i read your post many many times and read the question many times and tried to answer it by myself, but i just dont get it. i really dont know what to do, im sooo confused. what extactly is velocity? how do you spot it from a question?

"Speed is distance over time, but this is obvious because of their units" time isnt defined is it? man this sucks.... i hate that i dont get it...

5. Sep 6, 2004

### Alkatran

The standard unit for time is the second: s
The standard unit for distance is the meter: m
The stand unit for speed is meters per second: m/s
Guess how you go from time (s) and distance (m) to speed (or velocity) (m/s)

Obivously, distance/time (m/s) = speed (m/s)

6. Sep 6, 2004

### needhelpperson

Well i'll start you up.

You know that the motorcycle has to travel 27m + whatever distance the truck travels during that time.

since constant velocity. d = vt

well we know that initial velocity, so of the 5 kinematics equations we should use vi*t + .5*a*t^2 = distance traveled by motorcycle

I'm sure you'll know what to do with this information...

7. Sep 6, 2004

### Alkatran

vi*t + .5a*t^2 = vf
You have a, and the relation between vi and vf.
This means you need a second equation (two unknowns)

I can understand why this would be confusing. You can't just pick an equation here and plug it in. You need to pick two equations, plug them in to the data, then plug them into each other. That will solve on variable... and after that the whole thing falls into place.

8. Sep 7, 2004

### needhelpperson

uhh... its not confusing at all

since they both have same initial velocity

27 + vt = vt + .5at^2

wow the vt cancels out... WAY SIMPLER... As compared to this WOW...

Last edited: Sep 7, 2004
9. Sep 7, 2004

### Alkatran

That wasn't for this question, it was a general technique I use. It works wonders for me, so why not explain it? By the way:

27 + vt = vt + .5at^2
25 = .5at^2

You still have two variables, you still need a second equation. Where's the improvement? Why are you showing him a trick that will work in very few questions?

10. Sep 7, 2004

### needhelpperson

this is just to get him/her started, as i said before... I was hoping he/she can think of a way to use this information... And where's the improvement, well, i didn't see him going ne where so i might as well give him a little bit more hint...
And anyways, even if you explain your "technique", not everyone can understand it and take full advantage of it as easily.
Sometimes even though you had a good intention, like now in helping people, you might actually confuse them by accident. I would just let them discover their own techniques as they do more questions.

Last edited: Sep 7, 2004
11. Sep 7, 2004

### Alkatran

It looked like he just didn't know velocity is another word for speed, and confused the absolute time we oh-so-love to use in classic physics with relativity's time.

12. Sep 7, 2004

### needhelpperson

i think he's confused as to where to start, since even a simple thing like the velocity is bothering him....

Last edited: Sep 7, 2004
13. Sep 7, 2004

### EaGlE

your right, i never knew that velocity was speed. lol. it makes things wayyy easier. ok i found A.) it's 3 secs

ok working on B now. i cant seen to find the initial speed of the car.

using the formula. v(t) = v(0) + at

v(t) = 0 + (6m/s^2)(3)
so... v(t) = 18m/s^2 is the speed of the motocycle

and the car is twice as much right? 18*2 = 36 m/s

is that right?

14. Sep 7, 2004

### Alkatran

The bolded parts are corrected. Velocity (or speed) can't have a unit of m/s^2. (I didn't check the values... a bit late)

15. Sep 7, 2004

### EaGlE

(i know for sure the answer for A.) is 3secs)

actually, the motocyle has a two times the velocity of the car.
so would i divide 18/2 = 9. units dont matter at this online submission, but it's good to know. 36 was the wrong answer. i only get 2 tries left on this question, so im asking before i submit my answer agian.

Last edited: Sep 7, 2004