Physics Homework
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1.) A particle at t 1 = -2.0 s is at x 1 = 3.4 cm and at t 2 = 4.5 s is at x2 = 8.5 cm. What is its average velocity? Can you calculate its average speed from this data?
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2.) A horse canters away from its trainer in a straight line, moving 160 m away in 17.0 s. It then turns abruptly and gallops halfway back in 6.8 s. Calculate:
(a) Its average speed
(b) Its average velocity for the entire trip, using “away from the trainer”, as the positive direction.
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3.) A car traveling 90 km/h is 100 m behind a truck traveling 75 km/h. How
long will it take the car to reach the truck?
4.) At highway speeds, a particular automobile is capable of an acceleration of about 1.6 m/s2. At this rate, how long does it take to accelerate from 80 km/h to 110 km/h?
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5.)Determine the stopping distances for an automobile with an initial speed of 90 km/h and human reaction time of 1.0 s:
(a) For an acceleration a = -4.0 m/s2;
(b) For a = -8.0 m/s2.
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6.) Calculate:
(a) How long it took King Kong to fall straight down from the top of the empire state building (380 m high)
(b) His velocity just before “landing”.
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7.) A kangaroo jumps to a vertical height of 2.55 m. How long was it in the air before returning to Earth? 0.26 s.
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8.) A helicopter is ascending vertically with a speed of 5.60 m / s. At a height of 115 m above the earth, a package is dropped from a window. How much time does it take for the package to reach the ground?
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Can someone please provide me the answers and explanations to these problems because i am completely lost. I would appreciate any help. thanx =) I am trying me BEST to understand physics. It really isn't working.
I am not going to just do all of these for you. I will do the cantering horse since it seems to be typical.
It goes +160 m in the first 17 seconds
Then it goes - 80 m in 6.8 seconds
so the distance it traveled was 160 + 80 = 240 m in 17+6.8 = 23.8 s
average SPEED = 240/23.8 m/s
Now its total displacement from the starting point is 160 - 80 = +80 m in those 23.8 seconds so its velocity is
+80/23.8 m/s
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Now that had no acceleration so I will do 8, the helicopter.
a = -9.8 m/s
v = Vo + a t = 5.60 - 9.8 t
h = Ho + Vo t + (1/2)at^2
so 0=115 + 5.6 t -4.9 t^2
that is a quadratic equation
t = [-5.6 +/- sqrt (31.4+2254)] / 9.8
= [ -5.6 + 47.8 ] / 9.8
= 4.31 seconds
By the way, for #5 look at my reply to Marina yesterday
THANX SOO MUCH!! YOU ARE AWESOME!!
Sure! I'll help you with your physics homework problems one by one. I'll provide the answers and explanations for each problem.
1.) To find the average velocity, we use the formula:
Average Velocity = (Change in displacement) / (Change in time)
Given:
t1 = -2.0 s
x1 = 3.4 cm
t2 = 4.5 s
x2 = 8.5 cm
To calculate the change in displacement:
Change in displacement = x2 - x1 = 8.5 cm - 3.4 cm = 5.1 cm
To calculate the change in time:
Change in time = t2 - t1 = 4.5 s - (-2.0 s) = 6.5 s
Plugging these values into the formula:
Average Velocity = (5.1 cm) / (6.5 s)
The units for average velocity will be cm/s.
To calculate average speed from this data, we use the formula:
Average Speed = (Total distance traveled) / (Total time taken)
Since distance is a scalar quantity, it doesn't depend on direction. So, the average speed is the magnitude of the average velocity.
Average Speed = |Average Velocity| = |(5.1 cm) / (6.5 s)|
Average Speed will also have units in cm/s.
2.) For part (a) of this problem, we need to calculate the average speed of the horse.
Average Speed = Total distance traveled / Total time taken
Given:
Distance traveled away from trainer = 160 m
Time taken = 17.0 s
Average Speed = (160 m) / (17.0 s)
The units for average speed will be m/s.
For part (b), we need to calculate the average velocity of the entire trip using "away from the trainer" as the positive direction.
Average Velocity = (Change in displacement) / (Total time taken)
To calculate the change in displacement:
Change in displacement = (160 m / 2) = 80 m (since the horse travels halfway back)
Total time taken = 17.0 s + 6.8 s = 23.8 s
Average Velocity = (80 m - 160 m) / (23.8 s)
The units for average velocity will be m/s with the direction being negative.
3.) To find the time it takes for the car to reach the truck, we need to calculate the time it takes for the car to travel a distance of 100 m at a speed of 90 km/h.
Given:
Speed of the car = 90 km/h = (90 km/h) * (1000 m/km) / (3600 s/h) ≈ 25 m/s
Distance between car and truck = 100 m
Time = Distance / Speed = 100 m / 25 m/s = 4 seconds
Therefore, it will take the car 4 seconds to reach the truck.
4.) To find the time it takes for the car to accelerate from 80 km/h to 110 km/h with an acceleration of 1.6 m/s^2, we use the following formula:
Time = Change in velocity / Acceleration
Given:
Initial velocity = 80 km/h = (80 km/h) * (1000 m/km) / (3600 s/h) ≈ 22.2 m/s
Final velocity = 110 km/h = (110 km/h) * (1000 m/km) / (3600 s/h) ≈ 30.6 m/s
Acceleration = 1.6 m/s^2
Change in velocity = Final velocity - Initial velocity = 30.6 m/s - 22.2 m/s = 8.4 m/s
Time = (8.4 m/s) / (1.6 m/s^2)
The units for time will be in seconds.
5.) For both parts (a) and (b) of this problem, we can use the following formula to calculate the stopping distance:
Stopping Distance = (Initial velocity)^2 / (2 * |Acceleration|)
Given:
Initial velocity = 90 km/h = (90 km/h) * (1000 m/km) / (3600 s/h) ≈ 25 m/s
Human reaction time = 1.0 s
Acceleration (a) = -4.0 m/s^2 for part (a), -8.0 m/s^2 for part (b)
To calculate stopping distance for part (a):
Stopping Distance = (25 m/s)^2 / (2 * |-4.0 m/s^2|)
To calculate stopping distance for part (b):
Stopping Distance = (25 m/s)^2 / (2 * |-8.0 m/s^2|)
The units for stopping distance will be meters.
6.) For part (a) of this problem, we can use the formula of motion under constant acceleration to calculate the time it took for King Kong to fall:
Distance = Initial velocity * time + (1/2) * acceleration * time^2
Given:
Height of the Empire State Building = 380 m
Acceleration due to gravity (g) ≈ 9.8 m/s^2
Using the formula and rearranging, we get:
380 m = 0 * t + (1/2) * 9.8 m/s^2 * t^2
Simplifying the equation and solving for time 't', we get:
t = √(2 * (380 m) / (9.8 m/s^2))
The units for time will be seconds.
For part (b), just before "landing," King Kong's velocity will be equal to his maximum velocity when falling. We can use the following formula to calculate that:
Final velocity = Initial velocity + acceleration * time
Given:
Initial velocity = 0 m/s (assuming King Kong starts from rest)
Acceleration due to gravity (g) ≈ 9.8 m/s^2
Time 't' calculated from part (a)
Final velocity = 0 m/s + 9.8 m/s^2 * (time 't' calculated from part (a))
The units for final velocity will be m/s.
7.) To find how long the kangaroo was in the air before returning to Earth, we need to calculate the total time for its vertical jump.
Given:
Maximum vertical height = 2.55 m
To calculate the total time, we can use the following formula:
Time = √(2 * height / g)
where 'g' is the acceleration due to gravity, approximately 9.8 m/s^2.
Time = √(2 * 2.55 m / 9.8 m/s^2)
The units for time will be seconds.
8.) To find the time it takes for the package to fall from a height of 115 m, we can use the formula of motion under constant acceleration:
Height = Initial velocity * time + (1/2) * acceleration * time^2
Given:
Initial velocity = 5.60 m/s
Height = 115 m
Acceleration due to gravity (g) ≈ 9.8 m/s^2
Using the formula and rearranging, we get:
115 m = 5.60 m/s * t + (1/2) * 9.8 m/s^2 * t^2
Simplifying the equation and solving for time 't', we get:
t^2 + (5.60 m/s) * t - (115 m) * 2 / (9.8 m/s^2) = 0
Solving this quadratic equation will give us the value of time 't'. The units for time will be seconds.
I hope this helps you understand how to solve each problem step-by-step. If you have any further questions, feel free to ask!