Physics: 1-D and 2-D Kinematics

[kingwinner]

1) A racing car starts into a circular portion of a Grand Prix course at 200km/h[E], travelling in a clockwise direction. By the time it is headed due south, its speed has increased to 240km/h.

1a) If this took 12.0s, find the average acceleration during the turn.

1b) ESTIMATE the radius of the curve.

 

[i am OK with question 1a, I got that the acceleration is 7.23m/s^2 [s39.8W] , but I don't have the faintest idea for question 1b, if I need the radius, I would need the displacement, in which I don't have...even worst, the direction and speed are both changing...]

 

2) A power boat travels down the a river from City A to City B, at full throttle. The trip takes 3.0h. The boat then heads back to City A, again at full throttle. This time, the trip takes 15h. With no gas left, the boat now drifts with a steady current back to City B. How long does the third trip take?

 

["at full throttle", does this mean slowing down, speeding up, or at constant velocity? I am not an expert about the mechanics of a power boat...

Also, only time is given, how can I figure out the time taken for the third trip?]

 

Can someone help me? I would appreciate! :lol:

[arkain101]

in questions like this you can assume it means constant velocity and so, include it as a part of your concluded answer.

 

Example, in consideration that the boat moved at a constant velocity in all directions of travel, the third trip would take _____

 

(the answer is in speed of the river, that is calculated by the time difference of each direction of travel. While heading to B the boat is going river speed+boat speed in reference to the distance from a to b on land. and in the opposite direction the speed of the water will remove the velocity of the boat in reference to the land. So in that is the speed of water.)

 

the boat trip takes 5 times longer to go upstream. So the water must be moving at a factor of such and such to create the difference. water speed = boat drift time. Inject some fake distances. I tried to work it out and had some probs lol,

[CraigD]

1) A racing car starts into a circular portion of a Grand Prix course at 200km/h[E], travelling in a clockwise direction. By the time it is headed due south, its speed has increased to 240km/h.

1a) If this took 12.0s, find the average acceleration during the turn.

1b) ESTIMATE the radius of the curve.

 

[i am OK with question 1a, I got that the acceleration is 7.23m/s^2

Right – you appear to understand velocity vectors :lol: .

but I don't have the faintest idea for question 1b, if I need the radius, I would need the displacement, in which I don't have...even worst, the direction and speed are both changing...]

You’ve been told that: 1) the course is circular; 2) the car changed its direction from due East to due South; – 1/4th of a turn – 3) its velocity from 200 km/hr to 240 km/hr (assuming a constant acceleration, an average velocity of 220 km/hr). Ask yourself: how far did the car go in 12 seconds; and what radius must a circle have for 1/4th of its circumference to be that length?

[kingwinner]

Hello,

 

1) "...average velocity of 220 km/hr"

But velocity has direction and is equal to displacement / time

displacement is the straight line vector connecting the two points, so it's not quite following the circumference...

 

2) Let vo=velocity at full throttle without current from City A to City B

c=velocity of current

d=displacement from City A to City B

t=time taken for the third trip

 

vo=(d/3) -c

 

-vo=(-d/15) -c

 

c=d/t

 

I got 3 equations with 4 unknowns, how can I solve it?

[arkain101]

3 hours one way.. 15 hours the other way.

 

if the distance was 45km it would mean it went 45km/15hrs= 3km/h upstream. and 45km/3hrs=15km/h down stream. The speed of the boat remained the same both ways on water, but the river speed shifted the boats d/t.

 

So what adds up to 15 and subtracts to 3. 9+6=15 9-6=3.

 

So either the boat or the water could be going 9km/h while the other is doing 6km/h. Point is according to the times and the fact the river will add a V in one direction and remove a V in the other direction from the boats constant V.

The speeds and distances are optional. The question of that problem is how fast is the river going. Because that will tell you the time it will take to float down stream.

So it will take 45/9= 5hrs or 45/6= 7.5 hrs.. depending on boat is doing 9 with water 6, or water 9 and boat 6.

 

 

 

I could have made this all complex.. but I forget the equations.. and given the lack of info in that problem I think this is the answer lol.

[CraigD]

1) "...average velocity of 220 km/hr"

But velocity has direction and is equal to displacement / time

displacement is the straight line vector connecting the two points, so it's not quite following the circumference...

The problem states that the car is traveling on “a circular portion of a … course”, so the car is following the circumference of a circle. Though the direction of the velocity and acceleration vectors change constantly, making the problem appear very complicated, the “approximate the radius” problem is actually just the simple problem I describe in post #3 – find the distance traveled, then find the radius of a circle with a circumference 4 times that. Recognizing this is the “trick” to the problem.

[CraigD]

2) Let vo=velocity at full throttle without current from City A to City B

c=velocity of current

d=displacement from City A to City B

t=time taken for the third trip

Good. Those are all the variable you need. You don’t need the part I italicized, though, so should probably remove it, to avoid confusion.

vo=(d/3) -c

vo is a constant, and should be independent of any of the other given constants. The boat’s speed without current doesn’t depend on the distance between City A and B, nor the speed of the current.

 

Try solving these equations:

d = (vo +c)*3 ;traveling downstream, velocity is vo +c, time is given as 3 hrs

d = (vo –c)*15 ;traveling upstream, velocity is vo – c, time is given as 15 hours

t = d / c ;traveling at the speed of the current, time is distance / velocity of the current.

using these steps:

1) Set the right sides of the first 2 equations equal, and solve for c in terms of vo.

2) Substitute either of the right sides of first 2 equations for d in the 3rd equation. d disappears from the equation.

3) Substitute the equation found of step 1 everywhere c appears in the equation found in step 2. c disappears.

4) Simply the equation found in step 3. Replace any vo/vo terms with 1. vo disapears.

 

So, even though you can’t determines vo, c, or d with the information given, you can find t :)

[kingwinner]

1b) The solution did it this way:

"Using 2-D vector addition and pythagorean theorem, v1+v2 = 86.79m/s[E50S]. Then,

substitute this (v1+v2) into the formula below to find the displacement:

displacement= [(initial velocity + final velocity) /2] * (time interval)

=[86.79m/s[E50S] /2] * (12s)

=520.74m[E50S]

 

A right-angled triangle can be formed:

r^2 + r^2 = 520.74^2

2r^2=520.74^2

r=368.22m

r=368m "

 

But we all know that the above formula can only be used when the acceleration is constant, but in this question, the acceleration isn't constant! Let assume the magnitude of the acceleration is constant! But the direction is changing at every instant, so the acceleration is also changing at every instant! Would this method of finding the displacement still work, then?

 

I don't get how they can use the formula while the acceleration isn't constant........

[CraigD]

1b) The solution did it this way:

…

But we all know that the above formula can only be used when the acceleration is constant, but in this question, the acceleration isn't constant! Let assume the magnitude of the acceleration is constant! But the direction is changing at every instant, so the acceleration is also changing at every instant! Would this method of finding the displacement still work, then?

 

I don't get how they can use the formula while the acceleration isn't constant........

I agree – this approach seems to “simplify” much more than my suggestion that we assume the average scalar velocity (magnitude) is the average of its initial and final values. The difference in results is dramatic – they get a radius of 368 meters, while I get one of 467. Their result implies an average velocity tangent to the circular track of 48 m/s =~ 174 km/hr, less than either the initial or final velocity!

 

Though both approaches involve some simplifying assumptions, so are estimates only, I think (actually, I can prove) mine is more accurate. Sometimes, text books don’t chose very good examples :)

[arkain101]

Wow, I just looked over my post and I am way wrong.. ah jeez.

 

Can someone display the answer with the correct formula?

[CraigD]

Can someone display the answer with the correct formula [for problem 2]?

OK. Though it’s traditional to have the question/thread starter do this to show they understand the solution, I think kingwinner has moved on to other work, so I’ll do it.

 

Starting with the original problem

A power boat travels down the a river from City A to City B, at full throttle. The trip takes 3.0h. The boat then heads back to City A, again at full throttle. This time, the trip takes 15h. With no gas left, the boat now drifts with a steady current back to City B. How long does the third trip take?

A useful equation relating distance (D), velocity (V), and time (T) is

D = V * T

 

From the information given, and the question asked, we can rewrite this equation 3 different ways, in terms of the velocity of the boat (VB) and of the current (VC), and the time the 3rd trip takes (T), using hours as time units:

[1] D = (VC +VB) * 3

[2] -D = (VC -VB) * 15

[3] D = VC * T

 

Multiplying both sides of [2] by -1 gives

[4] D = (VB -VC) * 15

 

Setting the right sides of [1] and [4] equal gives

[5] (VC +VB) * 3 = (VB -VC) * 15

 

Expanding and collecting like terms gives

[6] 18VC = 12VB

 

Dividing both sides by 12,

[7] VC = (2/3)VB

 

Setting the right sides of [3] and [1] equal (we could also have used [3] and [4]) gives

[8] VC * T = (VC +VB) * 3

 

Substituting the right side for [7] for every occurrence of the left side in [8] gives

[9] (2/3)VB * T = ((2/3)VB +VB) * 3

 

Expanding and simplifying gives

[10] (2/3)VB * T = 5VB

 

Dividing both sides by (2/3)VB and noting the time units gives the answer

[11] T = 15/2 = 7.5 hours

[arkain101]

by golly,

 

appreciated craig, I am amazed I got the answer with the method I used. It was later I realized it could only be 7.5, since if the current were faster and the boat were slower, the boat wouldnt return from point b.

 

I remember those vector equations now, from highschool.. sheesh.