Question

Just a car starts to accelerate from rest with acceleration 1.4 m/s. A bus moving with constant velocity of 12 m/s passes it in parallel lane.
(a) How long before the car overtakes the bus?
(b) In what velocity will the car than be going?
(c) How far have the car gone at this instant?

Answer 1

bus distance = 12 t
car distance = (1/2) a t^2 = 0.7 t^2
so when
12 t = 0.7 t^2
t = 12/0.7 = 17.1 seconds
v car = a t = 1.4 * 17.1 = 23.9 m/s
distances the same for car and bus = 12 t

Answer 2

after t seconds,
distance of car = 0.7t^2 m
distance of bus = 12t m

a) when is .7t^2 = 12t
.7t^2 - 12t = 0
t(.7t - 12) = 0
t = 0 or t = 12/.7 = 120/7 seconds , obviously rejecting t = 0

b) velocity of car = 1.4t
when t = 120/7 , v = 1.4(120/.7) = 240 m/s (absurd answer! that would be 864 km/h)

c) d = 12(120/7) = 1440/7 m or appr 205 m

What a weird question. That acceleration way out of line, other than
it should say m/s^2. It might have been the case of an publisher just
changing units such as ft/s into metric units to be more current, or else
just making up numbers without thinking.

Answer 3

b) velocity of car = 1.4t
when t = 120/7 , v = 1.4(120/.7) = 240 m/s (absurd answer! that would be 864 km/h)
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tilt, I think you mean 1.4 (120 / 7.0 )

Answer 4

Anonymous, glad you caught that .
Now the question makes sense, my bad