Problem:
As you walk to class with a constant speed of 1.60m/s , you are moving in a direction that is 20.8∘ north of east.
Part A
How much time does it take to change your displacement by 22.0m east?
Part B
How much time does it take to change your displacement by 31.0m north?
distance = rate * time :)
speed east = Ve = 1.60 * cos 20.8
speed north = Vn = 1.60 sin 20.8
time east = 22/Ve
time north = 31/Vn
Thank yiu very much Damon!
You can use X=Xo+Vot+1/2at^2
To solve this problem, we can use the basic principles of trigonometry and kinematics. Let's break down the problem into its components:
Part A:
To change your displacement by 22.0m east, we need to find the time taken. Since you are walking with a constant speed of 1.60m/s, we can use the equation:
Time = Distance / Speed
In this case, the distance is 22.0m, and the speed is 1.60m/s. Plugging these values into the equation, we get:
Time = 22.0m / 1.60m/s
Simplifying the equation gives us:
Time = 13.75 seconds (rounded to two decimal places)
So, it takes approximately 13.75 seconds to change your displacement by 22.0m east.
Part B:
To change your displacement by 31.0m north, we need to find the time taken. Since you are walking with a constant speed of 1.60m/s, we can again use the equation:
Time = Distance / Speed
In this case, the distance is 31.0m, and the speed is 1.60m/s. Plugging these values into the equation, we get:
Time = 31.0m / 1.60m/s
Simplifying the equation gives us:
Time = 19.38 seconds (rounded to two decimal places)
So, it takes approximately 19.38 seconds to change your displacement by 31.0m north.
Remember, when solving problems like these, it's important to always check that the units are consistent throughout the calculations.