Hi, please help me solve these questions for my class. I have so much homework
A car traveling at 17 m/s accelerates at 3.3 m/s2 for 16 seconds. To the nearest meter how far does it travel?
To the nearest tenth of a m/s, what is the final velocity of the car in the previous problem?
A passenger in a helicopter traveling upwards at 13 m/s accidentally drops a package out the window. If it takes 21 seconds to reach the ground, how high to the nearest meter was the helicopter when the package was dropped?
To the nearest meter what was the maximum height of the package above the ground in the previous problem?
Sure, I can help you solve these questions. Let's break them down step by step:
1) To find how far the car travels, we can use the formula s = ut + 0.5at^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time taken.
Given:
Initial velocity (u) = 17 m/s
Acceleration (a) = 3.3 m/s^2
Time (t) = 16 seconds
Using the formula, we substitute the values:
s = (17 * 16) + (0.5 * 3.3 * 16^2)
s = 272 + 0.5 * 3.3 * 256
s = 272 + 422.4
s = 694.4 meters
Therefore, the car travels approximately 694 meters.
2) To find the final velocity of the car, we can use the formula v = u + at, where v is the final velocity.
Given:
Initial velocity (u) = 17 m/s
Acceleration (a) = 3.3 m/s^2
Time (t) = 16 seconds
Using the formula, we substitute the values:
v = 17 + (3.3 * 16)
v = 17 + 52.8
v = 69.8 m/s
Therefore, the final velocity of the car is approximately 69.8 m/s.
3) To find the height of the helicopter when the package was dropped, we can use the formula h = ut + 0.5gt^2, where h is the height, u is the initial velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken.
Given:
Initial velocity (u) = 13 m/s
Acceleration due to gravity (g) = 9.8 m/s^2
Time (t) = 21 seconds
Using the formula, we substitute the values:
h = (13 * 21) + (0.5 * 9.8 * 21^2)
h = 273 + (0.5 * 9.8 * 441)
h = 273 + 0.5 * 9.8 * 441
h = 273 + 0.5 * 4314.6
h = 273 + 2157.3
h = 2430.3 meters
Therefore, the helicopter was approximately 2430 meters high when the package was dropped.
4) To find the maximum height of the package above the ground, we can use the formula h = ut - 0.5gt^2, where h is the height, u is the initial velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken.
Given:
Initial velocity (u) = 13 m/s
Acceleration due to gravity (g) = 9.8 m/s^2
Time (t) = 21 seconds
Using the formula, we substitute the values:
h = (13 * 21) - (0.5 * 9.8 * 21^2)
h = 273 - (0.5 * 9.8 * 441)
h = 273 - 0.5 * 9.8 * 441
h = 273 - 0.5 * 4314.6
h = 273 - 2157.3
h = -1884.3 meters
Therefore, the maximum height of the package above the ground is approximately -1884 meters. Note that the negative sign indicates that the package has fallen below ground level.
I hope this helps you with your homework! Let me know if you have any further questions.