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?
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.
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.
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.
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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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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
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