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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 is the constant acceleration of the train?

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#2

(Original post by

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?

**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?

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#3

I suggest u write down what info we've been given. First of all, we know for sure the

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

Using this fact, we can use the quantities

After that, you will have to set up two equations and solve them simultaneously to find

Hope this helps, Best of Luck

**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:

Doing the same for the second car, but this time using s = ut+0.5at^2 the eqn becomes

Solve these simultaneously for

Remember that u2 = v1 = x

**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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(Original post by

I suggest u write down what info we've been given. First of all, we know for sure the

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

Using this fact, we can use the quantities

After that, you will have to set up two equations and solve them simultaneously to find

Hope this helps, Best of Luck

**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:

Doing the same for the second car, but this time using s = ut+0.5at^2 the eqn becomes

Solve these simultaneously for

Remember that u2 = v1 = x

**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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(Original post by

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

**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

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#6

**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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**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

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(Original post by

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.

**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.

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#9

(Original post by

Nvm you were right you just have to sub initial velocity of second car as x + 1.5a

**MWM2000**)Nvm you were right you just have to sub initial velocity of second car as x + 1.5a

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