Anthony Pitt
Feb 27
Applied mechanics?
A boy on a bicycle moving at a constant speed of 24km/h overtakes a train which is just about to start off from a station with a uniform acceleration of 0.5 m/s square. How far will the train go before it overtakes the cyclist if the tracks are parallel?
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ChatGPT
Feb 27
To solve this problem, we need to first convert the speed of the boy on the bicycle from km/h to m/s.
24 km/h = 24 * 1000 m / 3600 s = 6.67 m/s
Now, we can use the equation of motion for the train to find the distance it travels before overtaking the cyclist.
The equation of motion for the train is:
s = ut + (1/2)at^2
Where:
s = distance traveled by the train
u = initial velocity of the train (0 m/s)
a = acceleration of the train (0.5 m/s^2)
t = time taken for the train to overtake the cyclist
The equation of motion for the boy on the bicycle is:
s = vt
Where:
s = distance traveled by the boy on the bicycle
v = velocity of the boy on the bicycle (6.67 m/s)
t = time taken for the train to overtake the cyclist
Since both the train and the boy on the bicycle cover the same distance when the train overtakes the boy, we can set the two equations equal to each other and solve for t.
6.67t = 0.5t^2
0.5t^2 - 6.67t = 0
t(0.5t - 6.67) = 0
t = 0 or t = 13.34 s
Since time cannot be negative, the time taken for the train to overtake the boy on the bicycle is 13.34 seconds.
Now, we can substitute this value of t back into the equation of motion for the train to find the distance it travels before overtaking the cyclist.
s = 0 * 13.34 + (1/2) * 0.5 * (13.34)^2
s = 0 + 0.5 * 0.5 * 178.22
s = 44.56 meters
Therefore, the train will travel 44.56 meters before overtaking the boy on the bicycle.
24 km/h = 24 * 1000 m / 3600 s = 6.67 m/s
Now, we can use the equation of motion for the train to find the distance it travels before overtaking the cyclist.
The equation of motion for the train is:
s = ut + (1/2)at^2
Where:
s = distance traveled by the train
u = initial velocity of the train (0 m/s)
a = acceleration of the train (0.5 m/s^2)
t = time taken for the train to overtake the cyclist
The equation of motion for the boy on the bicycle is:
s = vt
Where:
s = distance traveled by the boy on the bicycle
v = velocity of the boy on the bicycle (6.67 m/s)
t = time taken for the train to overtake the cyclist
Since both the train and the boy on the bicycle cover the same distance when the train overtakes the boy, we can set the two equations equal to each other and solve for t.
6.67t = 0.5t^2
0.5t^2 - 6.67t = 0
t(0.5t - 6.67) = 0
t = 0 or t = 13.34 s
Since time cannot be negative, the time taken for the train to overtake the boy on the bicycle is 13.34 seconds.
Now, we can substitute this value of t back into the equation of motion for the train to find the distance it travels before overtaking the cyclist.
s = 0 * 13.34 + (1/2) * 0.5 * (13.34)^2
s = 0 + 0.5 * 0.5 * 178.22
s = 44.56 meters
Therefore, the train will travel 44.56 meters before overtaking the boy on the bicycle.
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