PHYSICS HELP - velocity, acceleration, etc.?

Question

PLEASE EXPLAIN BRIEFLY AND LABEL ANSWERS!

At the instant a traffic light turns green, a car starts from rest and accelerates uniformly at a rate of 4.0 metres/second^2 [E]. At the same instant, a truck travelling with a constant velocity of 90.0 km/hour [E] overtakes and passes the car.

a) How far beyond the starting point is the car after 10.0 seconds?

b) How far beyond the starting point is the truck after 10.0 seconds?

c) The car passes the truck at a distance of 312.5 metres beyond the starting point. How fast is the car travelling at this instant?

d) How long does the car take to pass the truck?

e) Both drivers suddenly see a barrier 100.0 metres away and hit their brakes at exactly the same time. Assuming that both vehicles decelerate uniformly, and they take 3.0 seconds to stop, will they stop in time?

Answer 1

a) remember the distance formula for constant acceleration motion? r = vt + (1/2)at^2, since the car start from start, v = 0. so we have r = (1/2)at^2 = 0.5 * 4 * 10^2 = 200 meters.
b) we know that r = vt for motions in constant velocity so
r = 90 * km / hour * 10 second = 90000/3600 * 10 = 250 meters.
c) again use the formula r = (1/2)at^2, we also have that v = at, using these two equations and eliminate t gives us r = (1/2)v^2/a
so v = squareroot of 2ar = 50 m/s
d ) it is the same time required to reach 50m/s for the car. v = at. so t = v/a = 50/4 = 12.5 sec.
e) now we need to use the more general formula V = v0 + at,
where V is the final velocity and v0 is the initial velocity, we know that the final velocity will be 0 if they could stop completely so v0 = -at, a = -v0/t, put this in the formula r = v0t+(1/2)at^2 gives you r = (1/2)v0t = 0.5 * 50 * 3 = 75 meter, so yes it only takes the car to run 75 meter before it stops. for the truck, the vecity of it after 12.5 sec is still 90km/hour or 25m/s, so r = 0.5 * 25 * 3 = 37.5 meter, so the truck will stop on time too.

Answer 2

no longer lots to describe. all and sundry of those could be formulae you have learnt at school or have on your textbook. a) t = d / v = a hundred / 25 = 4seconds b) a = (v - u) / t = (50 - 25) / 5 = 25 / 5 = 5ms^-2 c) s = ut + a million/2 at^2 = 25*5 + a million/2 * 5 * 5^2 = 187.5m d) d = vt = 50*5 = 250m

Answer 3

a) s = (1/2)at^2
s(10) = (1/2)(4)(10^2)
s(10) = 200 m

b) s = vt
s(10) = (90)(10) = 900 m

c) v^2 = 2as
v = √(2)(4)(312.5)
v = 50 m/s = 180 kph

d) v = at
t = 50/4
t = 12.5 s

e) v = at
v^2 = 2as = 2vs/t
s = vt/2
s(car) = 50*3/2 = 75 m
s(truck) = 25*3/2 = 37.5 m