waynejkruse10
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i have a set of question for some science homework, i have done 9 of them but i cant do the 10th. am i allowed to post it for some opinions?
it isnt an assesment
it isnt an assesment
homework question question
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waynejkruse10
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here we go:
A car 4.0m long came up behind a semi-trailer 20.0m long travelling at a stready 72km/h. After travelling behind the semi-trailer for some time the driver of the car decides to overtake. If the car pulled out from behind the semi-trailer with the front of the car 10.0m behind the rear of the trailer and accellerated at 3.5m/s/s (3.5ms^-2) until the rear of the car was 10.0m in front of the truck, how far would the car travel and how long would it take.
These questions involve the equations of motion:
v = u + at
v^2 = u^2 + 2as
s = ut + 1/2at^2
v - final velocity
u - initial velocity
a - accelleration
t - time
s - displacement (how far its moved)
So in this particular question:
a = 3.5ms^-2
u = 20ms^-1 (72km/hr)
s = ?
t = ?
v = ?
A car 4.0m long came up behind a semi-trailer 20.0m long travelling at a stready 72km/h. After travelling behind the semi-trailer for some time the driver of the car decides to overtake. If the car pulled out from behind the semi-trailer with the front of the car 10.0m behind the rear of the trailer and accellerated at 3.5m/s/s (3.5ms^-2) until the rear of the car was 10.0m in front of the truck, how far would the car travel and how long would it take.
These questions involve the equations of motion:
v = u + at
v^2 = u^2 + 2as
s = ut + 1/2at^2
v - final velocity
u - initial velocity
a - accelleration
t - time
s - displacement (how far its moved)
So in this particular question:
a = 3.5ms^-2
u = 20ms^-1 (72km/hr)
s = ?
t = ?
v = ?
waynejkruse10
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all the others were easy, just needed to substitute the values in, this makes me think.
Im doing the exact same stuff in my Year 10 Physics class.
Hmm..
Im thinking it cant rly be answered if we keep thinking of both the truck and the car travelling at 20m/s. Im not sure if u can do this but bear with me...
To work out the time-
Would it be the same to say both the truck and the car were at an initial velocity of 0. so lets say the car and the truck are stopped. the car being 10m behind the truck. it has to travel 44m.( the intial 10m +the end 10m+ the trucks length + one cars length).
So it travels 44m with and acceleration of 3.5m/s/s. We only need to find the time so we use the equation s = ut + 1/2at^2.
s=44m
u=0
t=?
a=3.5m/s/s
The time it takes turns out to be 5.014265364s (im assuming this is wrong cause its not a nice whole number but meh).
Now we have the time it took for the car to overtake the truck we can work out to total displacement. so we say its traveling at 20m/s- accelerating at 3.5m/s/s for 5.014265364s.
so to work out displacement we use the same equation as before - s = ut + 1/2at^2
s=?
u= 20m/s
t= 5.014265364s
a=3.5m/s/s
how far would the car travel and how long would it take?
Displacement = 144.2853073m Time taken= 5.014265364s
This was like a guess..im sure some1 knows the correct answer..
Hmm..
Im thinking it cant rly be answered if we keep thinking of both the truck and the car travelling at 20m/s. Im not sure if u can do this but bear with me...
To work out the time-
Would it be the same to say both the truck and the car were at an initial velocity of 0. so lets say the car and the truck are stopped. the car being 10m behind the truck. it has to travel 44m.( the intial 10m +the end 10m+ the trucks length + one cars length).
So it travels 44m with and acceleration of 3.5m/s/s. We only need to find the time so we use the equation s = ut + 1/2at^2.
s=44m
u=0
t=?
a=3.5m/s/s
The time it takes turns out to be 5.014265364s (im assuming this is wrong cause its not a nice whole number but meh).
Now we have the time it took for the car to overtake the truck we can work out to total displacement. so we say its traveling at 20m/s- accelerating at 3.5m/s/s for 5.014265364s.
so to work out displacement we use the same equation as before - s = ut + 1/2at^2
s=?
u= 20m/s
t= 5.014265364s
a=3.5m/s/s
how far would the car travel and how long would it take?
Displacement = 144.2853073m Time taken= 5.014265364s
This was like a guess..im sure some1 knows the correct answer..
I would split it up into two parts and set up a system of equations and find solve for the intersection of the two lines for each part.
You know enough to set up the third equation, s = ut + 1/2at^2, so it has 2 variables.
s=20t+1.75(t^2)-10 for the car. The 10 is there because its 10 behind the truck.
s=20t for the truck
Solve algebraically or plug into a graphing calculator to find the intersection between the two graphs. The ordered pair should tell you the time and distance until the front of the car is even with the back of the truck.
For the second half, use a similar method.
The distance the car needs to travel farther than the truck is 34m. (20m+10m+4m)
s=ut+1.75(t^2) For the car. You need to find the velocity of the car at the end of part 1 (v=u+at) using the t-value you got in the first part.
s=20t For the truck.
You need to find a spot on the car's graph 34m above the truck's graph.
I'm not entirely sure on this part and how to explain it, though. I'm pretty tired and I'm not sure if I did that right (I probably didn't). I don't have my calculator handy so I can't check it.
Add the answers from part 1 to part 2 to get the total time and distance.
PS. I'm sorry if my methods seem a little convoluted and hard to follow. I usually make up my own methods as I go along, taking an indirect route to the answer, so its kinda hard to explain to others. :beard:
I can't think any more tonight. I'll try to pick up where I left off in the morning. If anyone else has an easier method, then just ignore mine.
You know enough to set up the third equation, s = ut + 1/2at^2, so it has 2 variables.
s=20t+1.75(t^2)-10 for the car. The 10 is there because its 10 behind the truck.
s=20t for the truck
Solve algebraically or plug into a graphing calculator to find the intersection between the two graphs. The ordered pair should tell you the time and distance until the front of the car is even with the back of the truck.
For the second half, use a similar method.
The distance the car needs to travel farther than the truck is 34m. (20m+10m+4m)
s=ut+1.75(t^2) For the car. You need to find the velocity of the car at the end of part 1 (v=u+at) using the t-value you got in the first part.
s=20t For the truck.
You need to find a spot on the car's graph 34m above the truck's graph.
I'm not entirely sure on this part and how to explain it, though. I'm pretty tired and I'm not sure if I did that right (I probably didn't). I don't have my calculator handy so I can't check it.
Add the answers from part 1 to part 2 to get the total time and distance.
PS. I'm sorry if my methods seem a little convoluted and hard to follow. I usually make up my own methods as I go along, taking an indirect route to the answer, so its kinda hard to explain to others. :beard:
I can't think any more tonight. I'll try to pick up where I left off in the morning. If anyone else has an easier method, then just ignore mine.
waynejkruse10
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i read through your posts, then thought i could try it using Tckarky's method - it makes sense.
It doesnt matter what the initial velocity is and the final velocity is, all it matters is the accelleration. So if you can work out that the car takes around 5 seconds you can work out the final velocity.
If the car and the truck both start out at the same velocity, 20m/s, it doesnt matter what the initial velocity is as long as it is the same. And accelleration doesnt matter because all it signifies is that it increases in speed by 3.5m/s every second.
Thanks Guys.
Wayne
It doesnt matter what the initial velocity is and the final velocity is, all it matters is the accelleration. So if you can work out that the car takes around 5 seconds you can work out the final velocity.
If the car and the truck both start out at the same velocity, 20m/s, it doesnt matter what the initial velocity is as long as it is the same. And accelleration doesnt matter because all it signifies is that it increases in speed by 3.5m/s every second.
Thanks Guys.
Wayne
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