Question

27. A person is running at his maximum speed of 4 m/s to catch a train. When he is 6m from the
of the compartment the train starts to leave the station at a constant acceleration of Ims
how long it takes him to catch up the train​

Answer 1

Answer:

he can't catch the at any time

Explanation:

because his spead is very slow to that is required.

Answer 2

Answer:

time = 2 seconds

Explanation:

train

a = 1m/s/s

distance = Xt

time = t

$Xt= \frac{1}{2}at^{2}$.........(eq.1)

Man

distance Xm = X + 6m

velocity = 4m/s

Xm = V.t..........(eq.2)

Take note that the man will have to cover (6 + x)metres but they take the same time.

equating both equations,

.'. Xt = Xm

0.5at'2 = 4t - 6

$0.5t^{2} = 4t -6\\ finding the value of t\\\\2 (0.5t^{2} = 4t -6)\\t^{2} = 8t -12\\\\t^{2} -8t +12 = 0\\t = (8 +or- \sqrt{8^{2} - 4x1x12} )/ 2\\ \\t = (8 - 4)/2 \\ \\t = 2 seconds\\or\\\\ t = (8 + 4)/ 2\\t = 6 seconds$

but time must be 2 seconds because at t = 6 seconds the train would have already gone too far and the man would not catch it