Question

4. Let's say you're driving in your car, approaching a red light on the Camp Hill Bypass. A black Porsche is stopped at the light in the right lane, but there's no-one in the left lane, so you pull into the left lane. You're traveling at 40 km/hr, and when you're 15 meters from the stop line the light turns green. You sail through the green light at a constant speed of 40 km/hr and pass the Porsche, which accelerated from rest at a constant rate of 3 m/s2 starting at the moment the light turned green.
(a) How far from the stop line do you pass the Porsche?
(b) When does the Porsche pass you?
(c) If a Boston police officer happens to get you and the Porsche on the radar gun at the instant the Porsche passes you, will either of you be pulled over for speeding? Assume the speed limit is 50 km/hr.


5. A rock is shot up vertically upward from the edge of the top of the building. The rock reaches its maximum height 2 s after being shot. Them, after barely missing the edge of the building as it falls downward, the rock strikes the ground 8 s after it was launched. Find
(a) upward velocity the rock was shot at;
(b) the maximum height above the building the rock reaches; and
(c) how tall is the building?

Answer 1

40000/3600 = 11.11 m/s

15 m/11.11 m/s = 1.35 seconds to get to Porsche starting point

time t from porsche start.
Vp = 3 t
xp from light = (1/2)3 t^2

t car = (t - 1.35) seconds
xc from light = 11.11 (t-1.35)

when does xp = xc ?
1.5 t^2 = 11.11 (t-1.35)

1.5 t^2 - 11.11 t + 15 = 0

t = [ 11.11 +/-sqrt(123.5 - 90) ]/3
t = [ 11.11 +/- 5.79]/3
t = 1.77 or 5.63 seconds
the early time is when you pass the Porsche

at 5.63 seoonds the Porsche is going
3 (5.63) = 16.9 m/s
= *3.6 = 60.8 km/hr

Answer 2

v = Vi - 9.81 t
v = 0 at top
so
0 = Vi - 9.81*2
Vi = 19.6 m/s up

h = Vi t - 4.9 t^2
= 19.6(2) - 4.9(4)
= 19.6 meters above roof

then falls for 8 - 2 = 6 seconds from way up there
h+19.6 = 4.9 (36)
h = 157 meters, 50 stories high !

Answer 3

Thankyou so much Damon! I just confused where to put what numbers and what Equation to use.
Also to clarify:
Constant Speed is always = V1
Average Speed is V2=
Also how do you know when to use -9.8 m/s/s as constant acceleration and not 9.8 and also using 4.9 or -4.9 m/s/s?

Answer 4

g is -9.8 if up is positive

The only time I did not use - is when the thing was falling. I knew I had - on both sides. I could have written
- (h +19.6) = - 4.9 t^2
because it goes DOWN that distance with a negative acceleration starting at the tip top of its path.