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?
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
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 !
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?
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.
To solve each of these problems, we will need to apply the laws of motion and use relevant equations. Let's break down each problem and explain how to get the answers step by step.
4. Driving and Passing the Porsche:
(a) To find the distance from the stop line at which you pass the Porsche, we first need to determine the time it takes for you to reach that point after the light turns green.
The equation we'll use to calculate the time is:
Distance = Initial velocity × Time + (1/2) × Acceleration × Time^2
Let:
- Distance = 15 meters (distance from the stop line)
- Initial velocity = 40 km/hr = (40 x 1000) ÷ 3600 m/s (convert km/hr to m/s)
- Acceleration = 0 m/s^2 (since you maintain a constant speed)
- Time = unknown
Using the equation, we can rearrange it to solve for time:
Time = (2 × Distance) ÷ (Initial velocity + Final velocity)
In this case, Final velocity is the same as the initial velocity since you maintain a constant speed. Plug in the values and calculate the time.
(b) To determine when the Porsche passes you, we need to find the time it takes for the Porsche to reach your position from the moment the light turns green.
We can use the equation:
Distance = Initial velocity × Time + (1/2) × Acceleration × Time^2
Let:
- Distance = distance covered by the Porsche until it passes you
- Initial velocity = 0 m/s (since the Porsche starts from rest)
- Acceleration = 3 m/s^2
- Time = unknown
Rearrange the equation to solve for time and substitute the given values.
(c) To determine if you or the Porsche will be pulled over for speeding, we need to compare your velocities with the speed limit.
Convert the speed limit from km/hr to m/s and compare it to your velocity and the Porsche's velocity. If either of you exceeds the speed limit, you would be at risk of being pulled over.
5. Shooting Upwards and Falling Downward:
(a) To find the upward velocity with which the rock was shot, we need to use the equation for motion:
Final velocity = Initial velocity + (Acceleration × Time)
Let:
- Final velocity = 0 m/s (at maximum height, the velocity becomes zero)
- Acceleration due to gravity = -9.8 m/s^2 (negative because it acts against the upward direction)
- Time = 2 s (time taken to reach maximum height)
Rearrange the equation to solve for the initial (upward) velocity of the rock.
(b) To determine the maximum height above the building the rock reaches, we need to calculate the displacement during the upward motion.
Use the equation for displacement:
Displacement = Initial velocity × Time + (1/2) × Acceleration × Time^2
Let:
- Displacement = maximum height above the building
- Initial velocity = found in part (a)
- Acceleration = -9.8 m/s^2
- Time = 2 s
Substitute the values into the equation and solve for displacement.
(c) To find the height of the building, we need to calculate the total distance the rock travels from the top of the building to the ground.
Since the rock reaches the ground 8 seconds after being launched, we can use the equation for free fall motion:
Distance = Initial velocity × Time + (1/2) × Acceleration × Time^2
Let:
- Distance = height of the building
- Initial velocity = found in part (a)
- Acceleration = 9.8 m/s^2 (positive, as gravity is acting downwards)
- Time = 8 s
Substitute the values and solve for distance to find the height of the building.
Remember to calculate and convert all necessary units to solve each problem accurately.