Physics problem, exam tomorrow!

Jasonbot

macrumors 68020
Original poster
Aug 15, 2006
2,467
0
The Rainbow Nation RSA
Here's the problem:

A motor car A is at rest at a traffic light. as the light turns green A starts to accelerate and at the same moment a motorbike M moves past A with a constant velocity of 15m.s^-1.


Thats the Velocity Time graph, the values on the left and bottom are 15 and 10 respectively.

The question is, after how many seconds will car A catch Bike M.

I figured the acceleration of M=0 and A=1.5. and to work out the time I should use v*t=(v*t)/2. But then I get stuck...

Help appreciated, thanks!
 

Abstract

macrumors Penryn
Dec 27, 2002
24,414
124
Location Location Location
Basically, the question is asking: When do the bike and car travel the exact same distance (assuming distance d=0 at the traffic light)?


So, you want to use the equation:

d = v(i)^2 + 0.5a(t^2) for the car. v(i)= v(initial) = 0 m/s for the car.

and d = v(bike)t for the bike, since it doesn't accelerate.

so again, when do the car and bike travel the same distance?

d(car)= d(bike)

0.5at^2 = v(bike)t

Solve for t. ;)



------------------------------------------------------

If you want the easier way to solve it:

The time it takes for the car to have an "AVERAGE" speed of the bike is the time it takes for both vehicles to have travelled the same distance. So if the bike is travelling at a constant speed of 10 m/s, then how much time does it take for the car to have an average speed of 10 m/s? Well, the answer is: When the car accelerates and finally reaches a velocity of 20 m/s, it will have had an average velocity of 10 m/s over that timespan. So how long does it take for the car to reach a velocity of 20 m/s? ;)

-------------------------------------------

Conceptually: The car and bike have travelled the same distance when the area under the curve is the same for both the bike and the car. Look at the graph. It's a velocity vs. time graph. Look at the graph of the bike: How do you figure out the area of a rectangle in this case? It's just velocity x time. For the car, it's (velocity x time)/2.

When the two "curves" of the graph intersect, the area of the rectangle is 2x bigger than the area under the car's graph (ie: it's 2x bigger than the triangle).
 

swiftaw

macrumors 603
Jan 31, 2005
6,309
20
Omaha, NE, USA
Starting from rest, the distance covered in t seconds by an object accelerating at a constant a m/s/s is (a*t^2)/2

An object travelling at a constant b m/s covers a distance b*t in t seconds,


setting these two things equal to each other with a = 1.5 and b = 15, gives

1.5t^2 / 2 = 15t

thus

1.5* t^2 = 30t

t = 20 seconds
 

CanadaRAM

macrumors G5
Here's the problem:

A motor car A is at rest at a traffic light. as the light turns green A starts to accelerate and at the same moment a motorbike M moves past A with a constant velocity of 15m.s^-1.

The question is, after how many seconds will car A catch Bike M.

I figured the acceleration of M=0 and A=1.5. and to work out the time I should use v*t=(v*t)/2. But then I get stuck...

Help appreciated, thanks!
Since distance of the car = 1/2 * (Vi + Vf) * t and you know that Vi is zero, you just need to find out how long it takes the car to accelerate to 30 m/s

At that point, it will have covered the same distance as the motorbike

15 = (0+30)/2
 

swiftaw

macrumors 603
Jan 31, 2005
6,309
20
Omaha, NE, USA
Since distance of the car = 1/2 * (Vi + Vf) * t and you know that Vi is zero, you just need to find out how long it takes the car to accelerate to 30 m/s

At that point, it will have covered the same distance as the motorbike

15 = (0+30)/2
It takes 20 seconds for the car to get to 30 m/s since from the graph the acceleration is 1.5m/s/s
 

Jasonbot

macrumors 68020
Original poster
Aug 15, 2006
2,467
0
The Rainbow Nation RSA
Ah, What I was doing was going...

s=.5(1.5)t^2
s+75=15t
and I ended up with (t-10)^2 thus making t=10. I messed everything up. teh solution seems so much more logical now!

I just need to remember my equations... s=v.t and s=u.t+.5(a)(t^2):D


Oh yeah, 96.3% here I come...

EDIT: I was adding the 75 'coz another example used a boy chasing a train that started x meters ahead. sadly these 2x objects started together and thats why I stuffed up!
 

-hh

macrumors 68030
Jul 17, 2001
2,527
323
NJ Highlands, Earth
Starting from rest, the distance covered in t seconds by an object accelerating at a constant a m/s/s is (a*t^2)/2

An object travelling at a constant b m/s covers a distance b*t in t seconds,
These are both examples of the general equasion:

d = d(zero) + v(zero)t + (1/2)at^2.

What we know is that at the point where the automobile will pass the motorcycle, the two have travelled the same distance from the stoplight.

Thus, d(auto) = d(motorycle).

And the "show all work" form becomes:


d(auto-zero) + v(auto-zero)*t + (1/2)a(auto)*t^2 = d = d(mc-zero) + v(mc-zero)*t + (1/2)a(mc)*t^2

Known:

d(auto-zero) = zero, because we are the ones establishing our coordinates.
d(mc-zero) = zero (same reason)

v(auto-zero) = zero (as per instructions)
v(mc-zero) = 15 m/sec (as per instructions)

a(auto) = 1.5 m/sec^2 (as per instructions)
a(mc) = 0 m/sec^2 (as per instructions)

t = unknown

gives

1.5t^2 / 2 = 15t
Yup, which then in the "showing all work" mode, you then divide both sides by t:

(1.5/2)t^2/t = 15t/t

...and then multiply by 2 (because its more obvious this way):

1.5t = 15*2

...and divide by 1.5:

t = (30)/(1.5)


t = 20 seconds
Same answer I got.


FWIW, the first general formula listed above, namely:

d = d(zero) + v(zero)t + (1/2)at^2.

...is the second derivative of distance with respect to time. If you've had Calculus already, this is an "oh yeah" and you can see how to derive it. Otherwise, you'll just have to memorize it.

The reason I mention this is because of this physics formula which should also look familiar:

v = v(zero) + at

This is the first derivative of distance with respect to time, and it will also probably show up on a Physics exam problem.


In general, the 'trick' to solving these problems is to keep yourself organized by laying out what data you need to solve whichever formula, then how the facts that you know will fit together to allow you to solve simultaneous equasions.

For example, Part II of the above question could be to ask:

"...and what was the velocity of the automobile at the time that it catches up with the motorcycle?"

The answer here is pretty simple: its "v = v(zero) + at" when t = 20sec.

Similarly, Part III could have been:

"...and what was the distance at which the automobile catches up with the motorcycle?"

Answer here is to plug t=20 into the d = d0 + V0t + 1/2at^2 formula.

Be aware that really clever tests will ask you just this second or third part and not give you the hint as to what you need to solve for first (here, we used time). Just lay out the facts in an organized fashion and you'll figure out where there's gaps that either have to be solved for, or for which you need to figure out how you can mathematically eliminate them.


-hh
 

CanadaRAM

macrumors G5
I can't really make out the graph,

but assuming this is all taking place in a vacuum, if A=1.5, seems to me like it should be right about 8 sec.
Umm,,, what would being in a vacuum have to do with it? Acceleration is the change in velocity -- net of any effects like drag.

8 sec is badly off... think about it -- after 8 sec, the car is doing 12 m/s and has travelled 6 m. At this point the motorbike is still pulling away...
 

swiftaw

macrumors 603
Jan 31, 2005
6,309
20
Omaha, NE, USA
Umm,,, what would being in a vacuum have to do with it? Acceleration is the change in velocity -- net of any effects like drag.

8 sec is badly off... think about it -- after 8 sec, the car is doing 12 m/s and has travelled 6 m. At this point the motorbike is still pulling away...
I think you mean the car has travelled 48m after 8 seconds (0.5*1.5*8^2), whereas the bike has travelled 120m after 8 seconds (15*8)
 

Jasonbot

macrumors 68020
Original poster
Aug 15, 2006
2,467
0
The Rainbow Nation RSA
-hh said:
Mr. Physics himself has spoken! Thanks! I hope I do this in my exam tomorrow. My only problem will be knowing which formulae to use and when. I just hope that I don't stuff that up.

Thanks again EVERYONE, you guys are awesome. If you have cookies enabled please accept this as a token of my gratitude:

 

-hh

macrumors 68030
Jul 17, 2001
2,527
323
NJ Highlands, Earth
Mr. Physics himself has spoken! Thanks! I hope I do this in my exam tomorrow. My only problem will be knowing which formulae to use and when. I just hope that I don't stuff that up.

Thanks, but I will add one bit of correction to what I wrote above:

d(auto-zero) = zero, because we are the ones establishing our coordinates.
d(mc-zero) = zero (same reason)


My rationale (above) is misleading. Yes, we can choose our coordinate system, but what I missed pointing out was that it was stated that the motorcycle was in allignement with the automobile just as the light turned green.

As such, we had an equality here as a "Given" for our problem's starting conditions, and I should have pointed this out more clearly.

I happened to choose "zero" to be the value for this equality because it was a convenient and intuitive coordinate system. I got away with it because this was a simple problem, with neither vehicle having a head start, etc.

For example, if we were to say that the cycle had a 42m head start, we would still need d(zero) values defined for both of them, and we would still be able to pick what we wanted the coordinate system to be.

The given "head start" would have been defined as:

d(zero-mc) = d(zero-auto) + 42m

And we could set up our coordinates to either say:

d(zero-mc) = 0
...thus, d(zero-auto) = 42m

Or:
d(zero-auto) = 0
...thus, d(zero-mc) = -42m

Plugging either choice into:

d(auto-zero) + v(auto-zero)*t + (1/2)a(auto)*t^2 = d = d(mc-zero) + v(mc-zero)*t + (1/2)a(mc)*t^2

...will lead to the same answer, so neither one is "wrong". Its just that one coordinate system choice might be easier to understand, and/or be easier to work with. Just remember to be careful with your +/- signs on such a problem :)


-hh