Solved these physics questions but the answers don't match

[random_soldier1337]

1. Question: Rachel walks 2.80 Km to a gym at 6.00 Km/h. Upon reaching, she walks back the same distance at 7.70 Km/h as the gym was closed. Find her average velocity.

Attempt: I find the total time by: (2.8/6)+(2.8/7.7) ~=0.830 h. Divide 5.6 by this to get approximately 6.74 Km/h.

First thing I am confused by is, why is the average velocity simply not zero since her displacement over the entire time period is zero? (I actually calculated this since the back of the book gave a finite answer) Second can someone please confirm this answer? The back of the book says 3.27 Km/h but I don't know how that would be achieved.

2. Question: A car moves 0 to 900 m, starting and ending at rest. Through 225 m, it's acceleration is 2.75 m/(s^2) and through the remainder, the acceleration is -0.750 m/(s^2). What is the total travel time of the car through the 900 m?

Attempt: For the 225 m, from x-x0 = v0t + 0.5a(t^2), 225 = 0.5(2.75)(t1^2) => t1 = sqrt(1800/11) ~= 12.8 s. Find initial velocity from this for the rest of 675 m as v = v0 + at => v = 2.75sqrt(1800/11). Then from the equation for distance, 675 = 2.75sqrt(1800/11)t2 - 0.5(0.75)(t2^2), which is a quadratic and I solve it to get either approximately 66.9 s or 26.9 s neither of which help since the solution is around 55.2 (actually I had to use a solution book for this which uses slightly different values and gives 56.6 s as an answer, FYI). However, when I assume the 675 m to be covered in reverse so that the equation formulated is basically in reverse i.e. 675 = 0.5(0.75)(t2^2) this gives me t2 = sqrt(1800) ~=42.4 s, so that total time is 55.2 s.

So you can see why I am confused since I actually used a hunch to arrive at the answer and don't actually know what the correct method is i.e. consider the motion of the 675 m in forward with a finite initial velocity or in reverse with zero initial velocity?

3. Question: A train traveling at 161 Km/h, rounds a bend and the engineer is shocked to a see a locomotive 676 m ahead traveling in the same direction, on the same track at 29.0 Km/h. The engineer applies the brakes immediately. What must the constant deceleration be if the collision is to just be avoided?

Attempt: 161 Km/h = 805/18 m/s and 29 Km/h = 145/18 m/s. Dl = Dt + 676, where Dl = distance covered by locomotive over time to near collision and Dt = distance covered by train over time to near collision. Replacing values according to the formula for distance covered with constant acceleration, I get the equation, (145/18)t = (805/18)t - 0.5a(t^2) + 676. Since I need to find the acceleration I assume that the train wants to reduce speed to 0 m/s and from v = v0 + at => 0 = (805/18) + at => t = -(805/18a). Then replacing this value for t in the earlier equation, I basically solve for a and get a ~= 3.91 m/(s^2).

Seems reasonable but it is not the answer according to the answer book and when I replace this in the first equation to get a quadratic equation to help me find time, I get t ~= 30.2 s but replacing this in the original quadratic, the sides are unequal.

I have tried checking where I went wrong but I cannot find the answer, not to mention that I did not have much of an idea of how to proceed with this question to being with.

[Country Boy]

1. Question: Rachel walks 2.80 Km to a gym at 6.00 Km/h. Upon reaching, she walks back the same distance at 7.70 Km/h as the gym was closed. Find her average velocity.

 

Attempt: I find the total time by: (2.8/6)+(2.8/7.7) ~=0.830 h. Divide 5.6 by this to get approximately 6.74 Km/h.

​

First thing I am confused by is, why is the average velocity simply not zero since her displacement over the entire time period is zero? (I actually calculated this since the back of the book gave a finite answer) Second can someone please confirm this answer?

The average velocity is 0. Your book should have said "average speed".

 

The back of the book says 3.27 Km/h but I don't know how that would be achieved

An "average" of two numbers (the "average" here is not the arithmetic average but is an average) is always between the two numbers. Since 3.27 is less than both 6 and 7.7, that is impossible.

2. Question: A car moves 0 to 900 m, starting and ending at rest. Through 225 m, it's acceleration is 2.75 m/(s^2) and through the remainder, the acceleration is -0.750 m/(s^2). What is the total travel time of the car through the 900 m?

​

 

Attempt: For the 225 m, from x-x0 = v0t + 0.5a(t^2), 225 = 0.5(2.75)(t1^2) => t1 = sqrt(1800/11) ~= 12.8 s. Find initial velocity from this for the rest of 675 m as v = v0 + at => v = 2.75sqrt(1800/11). Then from the equation for distance, 675 = 2.75sqrt(1800/11)t2 - 0.5(0.75)(t2^2), which is a quadratic and I solve it to get either approximately 66.9 s or 26.9 s neither of which help since the solution is around 55.2 (actually I had to use a solution book for this which uses slightly different values and gives 56.6 s as an answer, FYI). However, when I assume the 675 m to be covered in reverse so that the equation formulated is basically in reverse i.e. 675 = 0.5(0.75)(t2^2) this gives me t2 = sqrt(1800) ~=42.4 s, so that total time is 55.2 s.

 

So you can see why I am confused since I actually used a hunch to arrive at the answer and don't actually know what the correct method is i.e. consider the motion of the 675 m in forward with a finite initial velocity or in reverse with zero initial velocity?

Your concept is correct. With initial velocity v0 and constant acceleration a, the distance traveled in time t is indeed (1/2)at^2+ v0t. Starting from rest the car goes (2.75/2)t^2= 1.375t^2 meters in t seconds. So it does the first 225 meter is sqrt(225/1.375)= 12.8 seconds, as you say, and its speed at that point is 2.75(12.8)= 35.2 m/s. That is again what you have. The distance the rest of the way, in t seconds is given by (-0.75/2)t^2+ 35.3t meters. To complete the 900- 225= 675 m requires (-0.375)t^2+ 35.3t= 675 or (0.375)t^2- 35.3t+ 675= 0. The quadratic formula gives two positive answers, 50.6 sec or 20.0 seconds.

3. Question: A train traveling at 161 Km/h, rounds a bend and the engineer is shocked to a see a locomotive 676 m ahead traveling in the same direction, on the same track at 29.0 Km/h. The engineer applies the brakes immediately. What must the constant deceleration be if the collision is to just be avoided?

 

Attempt: 161 Km/h = 805/18 m/s and 29 Km/h = 145/18 m/s. Dl = Dt + 676, where Dl = distance covered by locomotive over time to near collision and Dt = distance covered by train over time to near collision. Replacing values according to the formula for distance covered with constant acceleration, I get the equation, (145/18)t = (805/18)t - 0.5a(t^2) + 676. Since I need to find the acceleration I assume that the train wants to reduce speed to 0 m/s and from v = v0 + at => 0 = (805/18) + at => t = -(805/18a). Then replacing this value for t in the earlier equation, I basically solve for a and get a ~= 3.91 m/(s^2).

 

Seems reasonable but it is not the answer according to the answer book and when I replace this in the first equation to get a quadratic equation to help me find time, I get t ~= 30.2 s but replacing this in the original quadratic, the sides are unequal.

 

I have tried checking where I went wrong but I cannot find the answer, not to mention that I did not have much of an idea of how to proceed with this question to being with.

[random_soldier1337]

 

1. Question: Rachel walks 2.80 Km to a gym at 6.00 Km/h. Upon reaching, she walks back the same distance at 7.70 Km/h as the gym was closed. Find her average velocity.

 

Attempt: I find the total time by: (2.8/6)+(2.8/7.7) ~=0.830 h. Divide 5.6 by this to get approximately 6.74 Km/h.

​

First thing I am confused by is, why is the average velocity simply not zero since her displacement over the entire time period is zero? (I actually calculated this since the back of the book gave a finite answer) Second can someone please confirm this answer?

The average velocity is 0. Your book should have said "average speed".

 

The back of the book says 3.27 Km/h but I don't know how that would be achieved

An "average" of two numbers (the "average" here is not the arithmetic average but is an average) is always between the two numbers. Since 3.27 is less than both 6 and 7.7, that is impossible.

2. Question: A car moves 0 to 900 m, starting and ending at rest. Through 225 m, it's acceleration is 2.75 m/(s^2) and through the remainder, the acceleration is -0.750 m/(s^2). What is the total travel time of the car through the 900 m?

​

 

Attempt: For the 225 m, from x-x0 = v0t + 0.5a(t^2), 225 = 0.5(2.75)(t1^2) => t1 = sqrt(1800/11) ~= 12.8 s. Find initial velocity from this for the rest of 675 m as v = v0 + at => v = 2.75sqrt(1800/11). Then from the equation for distance, 675 = 2.75sqrt(1800/11)t2 - 0.5(0.75)(t2^2), which is a quadratic and I solve it to get either approximately 66.9 s or 26.9 s neither of which help since the solution is around 55.2 (actually I had to use a solution book for this which uses slightly different values and gives 56.6 s as an answer, FYI). However, when I assume the 675 m to be covered in reverse so that the equation formulated is basically in reverse i.e. 675 = 0.5(0.75)(t2^2) this gives me t2 = sqrt(1800) ~=42.4 s, so that total time is 55.2 s.

 

So you can see why I am confused since I actually used a hunch to arrive at the answer and don't actually know what the correct method is i.e. consider the motion of the 675 m in forward with a finite initial velocity or in reverse with zero initial velocity?

Your concept is correct. With initial velocity v0 and constant acceleration a, the distance traveled in time t is indeed (1/2)at^2+ v0t. Starting from rest the car goes (2.75/2)t^2= 1.375t^2 meters in t seconds. So it does the first 225 meter is sqrt(225/1.375)= 12.8 seconds, as you say, and its speed at that point is 2.75(12.8)= 35.2 m/s. That is again what you have. The distance the rest of the way, in t seconds is given by (-0.75/2)t^2+ 35.3t meters. To complete the 900- 225= 675 m requires (-0.375)t^2+ 35.3t= 675 or (0.375)t^2- 35.3t+ 675= 0. The quadratic formula gives two positive answers, 50.6 sec or 20.0 seconds.

3. Question: A train traveling at 161 Km/h, rounds a bend and the engineer is shocked to a see a locomotive 676 m ahead traveling in the same direction, on the same track at 29.0 Km/h. The engineer applies the brakes immediately. What must the constant deceleration be if the collision is to just be avoided?

 

Attempt: 161 Km/h = 805/18 m/s and 29 Km/h = 145/18 m/s. Dl = Dt + 676, where Dl = distance covered by locomotive over time to near collision and Dt = distance covered by train over time to near collision. Replacing values according to the formula for distance covered with constant acceleration, I get the equation, (145/18)t = (805/18)t - 0.5a(t^2) + 676. Since I need to find the acceleration I assume that the train wants to reduce speed to 0 m/s and from v = v0 + at => 0 = (805/18) + at => t = -(805/18a). Then replacing this value for t in the earlier equation, I basically solve for a and get a ~= 3.91 m/(s^2).

 

Seems reasonable but it is not the answer according to the answer book and when I replace this in the first equation to get a quadratic equation to help me find time, I get t ~= 30.2 s but replacing this in the original quadratic, the sides are unequal.

 

I have tried checking where I went wrong but I cannot find the answer, not to mention that I did not have much of an idea of how to proceed with this question to being with.

 

 

1. Oh right! I was right all along. Should have known that. Silly me!

 

2. I don't know how you are getting those values. I even checked the zeros of that same curve graphically with my TI-84 calculator and the zeros occur at roughly the same time as the values I had previously calculated by hand. According to the calculator they are 67.001661 (bad approximation since the y-value on the calc is 10^-10) and 26.865006.

[imatfaal]

 

1. Oh right! I was right all along. Should have known that. Silly me!

 

2. I don't know how you are getting those values. I even checked the zeros of that same curve graphically with my TI-84 calculator and the zeros occur at roughly the same time as the values I had previously calculated by hand. According to the calculator they are 67.001661 (bad approximation since the y-value on the calc is 10^-10) and 26.865006.

 

I also don't know where HofI is getting those results from

 

http://www.wolframalpha.com/input/?i=.5*.75*x%5E2-35.2x%2B675%3D0

 

I get the same as you.

 

IMPORTANT BIT

 

If you are getting two results that means the car is passing that point at two points in the future - but you are told it stops there! It is only there once - this is a clue that something is wrong IMHO.

 

I approached the problem using a different motion equation

 

v²=u²+2as

 

You have v₀ = 0 and v₂=0 and need to find v₁

You also know the distance traveled between t₀ and t₁ and between t₁ and t₂ -what is more you know the acceleration between these times

 

You can set up two equations

 

[latex]v_1^2=v_0^2+2 \cdot a_{01} \cdot s_{01}[/latex]

 

[latex]v_2^2=v_1^2+2 \cdot a_{12} \cdot s_{12}[/latex]

 

Both equations should solve out to give the same answer for v_1 - they do not. I think the question is wrong

[random_soldier1337]

@imatfaal,

 

Yeah, I believe the textbook itself has stated a wrong figure. If the assertion that the car ends at rest is taken, then it takes ~46.9 s after the first ~12.8 s to stop from the speed achieved during the first 225 m but it stops at 825 m. If it is assumed to stop at 675 m and either of the times is used, in the case of ~26.9 s, the car is not at rest (~15 m/s) at 675 m and if the time is taken as ~66.9 s, the final velocity is ~ -14 m/s which doesn't make any sense. Well at least for the method I used. Maybe one or the other figures given in the statement needs to be changed.

 

Anyway, thanks for the info.

 

I still don't know how to do the third one, though.

Edited January 17, 2017 by random_soldier1337

[imatfaal]

@imatfaal,

 

Yeah, I believe the textbook itself has stated a wrong figure. If the assertion that the car ends at rest is taken, then it takes ~46.9 s after the first ~12.8 s to stop from the speed achieved during the first 225 m but it stops at 825 m. If it is assumed to stop at 675 m and either of the times is used, in the case of ~26.9 s, the car is not at rest (~15 m/s) at 675 m and if the time is taken as ~66.9 s, the final velocity is ~ -14 m/s which doesn't make any sense. Well at least for the method I used. Maybe one or the other figures given in the statement needs to be changed.

 

Anyway, thanks for the info.

 

I still don't know how to do the third one, though.

 

The third question is quite funky - I am sure there is a straightforward way to solve this but I cannot see it.

 

I did it as follows

 

we know

v_fast_0 = 161km/h = 44.72m/s

v_slow_0= 29km/h= 8.06m/s

S_fast_0= 0 metres

S_slow_0=676 metres

 

we can set up equations of S as a function of time (t) and acceleration of the fast train (a_fast)

S_slow(t) = S_slow_0 + v_slow_0.t

S_slow(t) = 676 + 8.06.t

 

S_fast(t) = S_fast_0 + v_fast_0.t +1/2. a_fast.t^2

S_fast(t) = 0 + 44.72.t + 1/2.a_fast.t^2

 

we can then set the distance S for both trains to be equal to each other

676 + 8.06.t = 44.72.t + 1/2.a_fast.t^2

this is a quadratic in t - with a second unknown a_fast

 

1/2.a_fast.t^2 +(44.72-8.06)t - 676 = 0

 

Similar to the question 2 we only want a solution with a single point in time at which the two distances are equal - a quadratic has a single root when the discriminant equals zero.

 

so for ax^2+bx+c = 0 the discriminant is b^2-4ac - thus in our case we set

 

0 = (44.72 - 8.06)^2 - 4 . a_fast . (-676)

rearrange

4 . 1/2.a_fast . (-676) = (44.72 - 8.06)^2

a_fast = [(44.72 - 8.06)^2] / [4.1/2 (-676)] = -.994 m/s

 

I did a quick model on excel and this seems right - the trains are at 1cm apart at 37 seconds and at some point between 36 and 38 will be at 0m distance apart.

 

You notice that I did not solve the quadratic in t - just used the fact that I knew the discriminant must be zero to find out a_fast. I could then substitute in the value I have obtained for a_fast and solve for t to find the point of closest approach.

http://www.wolframalpha.com/input/?i=-1%2F2*((36+2%2F3)%5E2)%2F(4*.5*676)*t%5E2%2B36+2%2F3*t-676%3D0

[random_soldier1337]

@imatfaal,

 

That is indeed the correct answer. I just don't understand how the assumption for the single root could be made. With that assumption in most cases generally, the answer we may require (even if only one), would change by putting the discriminant as zero, if there are two roots, wouldn't it? How then could we guarantee that whether the assumption would be correct? It worked out here but how do we know generally?

 

I also don't understand how my method did not work. Isn't it implied that the acceleration should be large enough that the train should stop just before touching the locomotive? Then we could use the equation for final velocity to replace time in the quadratic and find the acceleration from that. I got -0.946 m/s^2 (so close yet so far!) and both of the sides are equal in the original distance equation the quadratic was derived from (approx. 1057 m).

[imatfaal]

@imatfaal,

 

That is indeed the correct answer. I just don't understand how the assumption for the single root could be made. With that assumption in most cases generally, the answer we may require (even if only one), would change by putting the discriminant as zero, if there are two roots, wouldn't it? How then could we guarantee that whether the assumption would be correct? It worked out here but how do we know generally?...

 

Think of a simple situation with s = ut +1/2at^2 and you will understand what the two or one solutions each mean.

 

1. You are traveling in positive x direction at S_0 metres with a velocity v m/s.

2. You start decelerating (ie accelerating in the -ve x direction) -a m/s^2.

3. Soon after you start decelerating you pass point Y but you are still traveling in +ve x direction

4. Some time later you come to a halt - and if the acceleration continues you start to move in the -ve x direction

5. Sooner or later you pass point Y again - this time traveling in the opposite ie -ve x direction

6. That is the situation you get with TWO roots for an S_Y = ut +1/2at^2 scenario

 

 

If we take the same first two points as above and posit a new scenario from 3 onwards

3a. Call the point Z where you actually stop (only for a briefest moment) before you start to accelerate back in the -ve x direction

4a. You never ever (with a constant acceleration) get back to Z

5a. This is now the situation with a SINGLE root for the S_Z = ut +1/2at^2 scenario

 

In both questions we were dealing with a "just got there" or "just stopped from crashing" These indicate you were at a turning point - ie the point of intersection was the minima (or maxima) of the parabola; and that is just a single point.

 

I will take a look at your solution and see what went wrong later

Attempt: 161 Km/h = 805/18 m/s and 29 Km/h = 145/18 m/s. Dl = Dt + 676, where Dl = distance covered by locomotive over time to near collision and Dt = distance covered by train over time to near collision. Replacing values according to the formula for distance covered with constant acceleration, I get the equation, (145/18)t = (805/18)t - 0.5a(t^2) + 676. Since I need to find the acceleration I assume that the train wants to reduce speed to 0 m/s and from v = v0 + at => 0 = (805/18) + at => t = -(805/18a). Then replacing this value for t in the earlier equation, I basically solve for a and get a ~= 3.91 m/(s^2).

OK - the reason this doesn't work is you do not need to decelerate to zero - just to 29 km/h

 

http://www.wolframalpha.com/input/?i=(145%2F18)(-660%2F(18a))+%3D+(805%2F18)(-660%2F(18a))+%2B+0.5a(-660%2F(18a))%5E2+%2B+676

 

That gets the same answer as me

 

That's a different approach from mine - I like mine better because the solution is easier maths ; but yours is probably the way it was intended to be solved

[random_soldier1337]

@imatfaal,

 

I think I understand. I just went for my method because I wasn't really thinking that far. Also just in case anyone else refers to this thread in the future, the 676 m in my initial equation was listed on the wrong side as well but other than that, the amendments made to the initial attempt get the correct answer.