# Physics question help

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#1
As a student arrives at Union station his train speeds off. The cars are 8.6m long; the first car takes 1.5s to pass the student whereas the following car takes 1.1s to pass.
What is the constant acceleration of the train?
0
11 months ago
#2
(Original post by MWM2000)
As a student arrives at Union station his train speeds off. The cars are 8.6m long; the first car takes 1.5s to pass the student whereas the following car takes 1.1s to pass.
What is the constant acceleration of the train?
what have you tried?
0
11 months ago
#3
I suggest u write down what info we've been given. First of all, we know for sure the s (dist/length) of the cars (8.6) and the t (time taken) (1.5s first car, 1.1s the second car) We then have to use SUVAT
We are trying to find acceleration, a
Please note that they haven't told us the initial velocity of the first car or that it's starting from rest, so we don't know u.
What we DO know (if I'm not wrong) is that the initial velocity, u2, of the second car is the same as the final velocity, v1, of the first car.
Using this fact, we can use the quantities s, t, x ( x is v1 of first car = u2 of second car) and a (which is same for both cars)
After that, you will have to set up two equations and solve them simultaneously to find a and x
Spoiler:
Show
The two equations must use the information we have. Since we know t=1.5 and s=8.6 for the first car, by using s= vt - 0.5at^2, we can say for the first car: 8.6 = x(1.5) - 0.5a(1.5^2)
Doing the same for the second car, but this time using s = ut+0.5at^2 the eqn becomes 8.6= x(1.1) + 0.5a (1.1^2)
Solve these simultaneously for a
Remember that u2 = v1 = x

Hope this helps, Best of Luck
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#4
(Original post by ZedEmDoubleU)
I suggest u write down what info we've been given. First of all, we know for sure the s (dist/length) of the cars (8.6) and the t (time taken) (1.5s first car, 1.1s the second car) We then have to use SUVAT
We are trying to find acceleration, a
Please note that they haven't told us the initial velocity of the first car or that it's starting from rest, so we don't know u.
What we DO know (if I'm not wrong) is that the initial velocity, u2, of the second car is the same as the final velocity, v1, of the first car.
Using this fact, we can use the quantities s, t, x ( x is v1 of first car = u2 of second car) and a (which is same for both cars)
After that, you will have to set up two equations and solve them simultaneously to find a and x
Spoiler:
Show
The two equations must use the information we have. Since we know t=1.5 and s=8.6 for the first car, by using s= vt - 0.5at^2, we can say for the first car: 8.6 = x(1.5) - 0.5a(1.5^2)
Doing the same for the second car, but this time using s = ut+0.5at^2 the eqn becomes 8.6= x(1.1) + 0.5a (1.1^2)
Solve these simultaneously for a
Remember that u2 = v1 = x

Hope this helps, Best of Luck
But that would give only one value of x and the simultaneous equations wouldn’t be correct as they imply that the initial velocity of car 1 and the initial velocity of car 2 are the same
0
#5
(Original post by MWM2000)
But that would give only one value of x and the simultaneous equations wouldn’t be correct as they imply that the initial velocity of car 1 and the initial velocity of car 2 are the same
I might be wrong though
0
11 months ago
#6
(Original post by MWM2000)
As a student arrives at Union station his train speeds off. The cars are 8.6m long; the first car takes 1.5s to pass the student whereas the following car takes 1.1s to pass.
What is the constant acceleration of the train?

You have two points, at which you know s and t:

The end of the first car: {s, t} = {8.6, 1.5}
The end of the second car (from the same initial point): {s, t} = {2*8.6, 1.5+1.1}

You can therefore get two equations with two unknowns (u and a), so can solve for them.
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#7
(Original post by MWM2000)
But that would give only one value of x and the simultaneous equations wouldn’t be correct as they imply that the initial velocity of car 1 and the initial velocity of car 2 are the same
Nvm you were right you just have to sub initial velocity of second car as x + 1.5a
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#8
(Original post by RogerOxon)

You have two points, at which you know s and t:

The end of the first car: {s, t} = {8.6, 1.5}
The end of the second car (from the same initial point): {s, t} = {2*8.6, 1.5+1.1}

You can therefore get two equations with two unknowns (u and a), so can solve for them.
Yes it works that way as well thankyou
0
11 months ago
#9
(Original post by MWM2000)
Nvm you were right you just have to sub initial velocity of second car as x + 1.5a
Yep, you got it
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