Nona Z.

asked • 10/03/16

A difficult question to understand in kinematics .

Tow trains , A and B are travelling on parallel tracks and in opposite directions . They start simultaneously from rest at stations a distance d m apart and head towards each other , with constant acceleration 0.4 m.s-2 and 0.2 m.s-2 respectively . After 50 s the fronts of the trains are level .
Calculate ,
A- the speed of train when they meet  .
B- the value of d .

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Arturo O. answered • 10/03/16

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Part A:
 
v = at
 
vA = aAt = (0.4)(50) m/s = 20 m/s
 
vB = aBt = (0.2)(50) m/s = 10 m/s
 
Part B:
 
Train A travels a distance
 
dA = (1/2)aAt2 = (1/2)(0.4)(502) m = 500 m 
 
Train B travels a distance
 
dB = (1/2)aBt2 = (1/2)(0.2)(502) m = 250 m
 
d = dA + dB = (500 + 250) m = 750 m
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Nona Z.

Thank you very much,
But my wondering is , when we calculated the speed when they met we used the acceleration for each one , but how can we use the t the same as both of them have different distances they've crossed to make , what if one of them has passed 30 seconds where the other one 20 seconds which gives us 50 second the total of the time , so if we can say that as long as they have different acceleration and different speeds and different distances so they should have different times if it makes sense to say that ? 
 
I can't stop thinking about this ,thak you very much 
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10/09/16

Arturo O.

They both began to travel at the same time and when they meet, they both arrive at the same point at the same time.  True, they  traveled different distances at different speeds.  But when they meet, they are both at the same place at the same instant in time.  The 50s is not the sum of their travel times.  They both traveled 50s to arrive at the same spot.  The sum of distances traveled during the 50s interval must equal the total initial separation distance.
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10/09/16

Nona Z.

Oh now I got the idea, so 50 s for each of them , that was brilliant thak you very much Sir .
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10/09/16

Arturo O.

You are welcome.  Feel free to post additional comments if you have more questions.
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10/09/16

Nona Z.

Thanks again,I have another exercise I'll feel glad if you helped me how to solve it,
 
A particle travelling in a straight line with acceleration a passes a point A whilst moving with velocity u . When it reaches a point B it has velocity v , and at this point its acceleration changes to -a .Show that when it again passes through A its speed is
square root of 2v2 _ u2
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10/09/16

Arturo O.

This may be hard to visualize without a picture, but here goes:
Assume you are moving left to right under constant acceleration a from point A to point B, with distance d1 from A to B, with initial speed u at A, and final speed v at B. From the laws of kinematics,

v2 = u2 + 2ad1

d1 = (v2 – u2 ) / (2a)

Now you are at point B, at speed v. You will overshoot point B and proceed at a new acceleration –a until the speed drops to zero at a point C to the right of point B, and then you reverse direction and return toward point A. Suppose the distance from B to C is d2. The initial speed from B to C is v at B, and the final speed is zero at C. Then from the same laws of kinematics used above,

02 = v2 + 2(-a)d2

d2 = v2 / (2a)

The total distance traveled from C back to A is d1 + d2. The acceleration from C to A is –a. We can compute the distance d1 + d2 from C to A from the magnitude of the acceleration. If it takes time t to get from C to A, then
 
(1/2)at2 = d1 + d2

t = [(2/a)(d1 + d2)]1/2
 
We want the final speed at A; call it vA.
 
vA = vC – at = 0 – at = -at (remember we reached speed zero at C before turning around)

vA = -a [(2/a)(d1 + d2)]1/2
 
Recall from above that

d1 = (v2 – u2 ) / (2a)

d2 = v2 / (2a)

Then
 
vA = = -a {(2/a) [(v2 – u2 ) / (2a) + v2 / (2a)]}1/2

Simplify this expression for vA and get

vA = -(2v2 – u2)1/2
 
The minus sign just means you are moving toward the left.
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10/09/16

Nona Z.

Oh thank you very much Mr.Arturo that was really helpful and clear I managed to understand everything and how it went .
Very clear explaining , Thank again Sir .
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10/09/16

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