Question

At 8:10am, train A leaves it's station heading east at 6m/s. At 8:15am, train B pulls away from its station and accelerates at 0.05m/s^2. Where does train B catch up to train A

Answer 1

Train B cacthes up to train A at a distance of 1953 m from the station

Explanation:

The motion of train A is a uniform motion at constant speed, so we can write the position of train A at time t as

$x_A(t) = v t$ (1)

where

$v = 6 m/s$ is the speed of train A

t is the time in seconds, measured starting from 8:10 am

Train B instead moves at constant acceleration, so the position of train B at time t is given by

$x_B(t) = \frac{1}{2}a(t-300)^2$

where

$a=0.05 m/s^2$ is the acceleration of train B

and where 300 is the time (in seconds) at which train B starts its motion after 8:10 am. In fact, train B leaves at 8:15 am, which means 5 minutes after train A, so 300 seconds after train A.

Train B catches up to train A when the two positions are equal:

$x_A = x_B\\vt = \frac{1}{2}a(t-300)^2$

Solving for t, we find:

$6t = \frac{1}{2}(6)(t-300)^2\\6t = 3t^2-1800t +270,000\\3t^2-1806t+270,000 = 0$

And solving the equation, we find two solutions:

t = 276.5 s

t = 325.5 s

However, we said that train B leaves the station only 300 seconds after train B: therefore, this means that t cannot be less than 300 s, so the correct solution is

t = 325.5 s.

Now we can find where the train B catches train A, by substituting this time into eq.(1):

$x_A = (6)(325.5) = 1953 m$

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