Speed, Velocity and Acceleration... Catching a Train Physics Help
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Speed, Velocity and Acceleration... Catching a Train Physics Help

[From: ] [author: ] [Date: 12-02-16] [Hit: ]
how would you set up the formulas to figure this out and graph it? They should intersect somewhere right?5*t = 1*t + ½(.1)*t² + 50 → .t = 15.t = 64.......
A man begins to catch up to a train 50 meters away. He is traveling at a constant: 5 m/s. The train is moving at 1 m/s and is accelerating .1 m/s². At what time will the man catch up to the train?

Also, how would you set up the formulas to figure this out and graph it? They should intersect somewhere right?

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Equate the distances of each:

Dm = Dt

5*t = 1*t + ½(.1)*t² + 50 → .05t² - 4t + 50 = 0

t = 15.5 sec
t = 64.5 sec

The second time is when he keeps running and the train catches up to him after he passes it.

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Place the origin at the man from the start.
So xm0 = 0 and xt0 = 50.
Now let's come up with the position function for both the man and the train.

vm = 5
xm = 5t

vt = .1t + 1
xt = .05t^2 + t + 50

These two equations (xm and xt) are the formulas you need to answer this problem.
Simply set them equal and solve for t. This value of t is when they are at the same position.

.05t^2 - 4t + 50 = 0
t^2 - 80t + 1000 = 0

This can be solved using the quadratic formula.

t = 40 +- 10sqr(6)
t = 15.5 s or 64.5 s

The graphs of the two position functions intersect at two places.
The man will first catch up to the train when t = 15.5 s and surpass the train
until t = 64.5 s when they meet again and the man starts to fall behind.

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You have typed .1 m/s^2 which had mislead me
The usual procedure is to add a zero and write 0.1 m/s^2.
Now noting it, I am deleting my previous answer and the correct answer is as follows

5t = 1t + 0.5*0.1*t^2 + 50

t = 15.5 s
and
t = 64.5 s

He catches the train at the time of 15.5 s
And after this time he is moving ahead of the train and hence he slows down to catch the train.
Between 15.5s and 64.5 s he will surely catch the train.
=========================

The straight line y = 5t intersects the parabola y =t + 0.05t^2 + 50 at two points
t 1 = 15.5 and t2 = 64.5
============================

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find the point when the distance covered by the train, (s) matches the distance covered by the man, (s+50).
for train: s = u +at
u = initial velocity
a = acceleration
t = time
for man: (s+50) = v*t
v = velocity
cancel out the s to get an equation in terms of time
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keywords: and,Acceleration,Speed,Train,Catching,Velocity,Physics,Help,Speed, Velocity and Acceleration... Catching a Train Physics Help
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