How Do You Solve These Physics Problems on Motion and Vectors?

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The discussion focuses on solving physics problems related to motion and vectors. The first problem involves calculating the velocity of a stone thrown with an initial velocity of 7.0 m/s and a constant acceleration of -9.8 m/s² after 0.50 seconds, with a participant confirming the answer as 2.1 m/s. The second problem requires determining the speed of an airplane accelerating from rest at 10.8 m/s² over a 400m runway, with guidance provided on using kinematic equations. The third problem involves finding the components of a displacement vector of 34.0m at an angle of 60.0 degrees east of north, with hints on using trigonometric functions to resolve the vector. Overall, participants seek clarification on applying kinematic equations and vector decomposition in these scenarios.
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1.) A stone is thrown with an initial velocity of 7.0m/s. the acceleration of the stone is constant and has the value of -9.8m/s to the power of ((2)). What is the velocity of the stone after .50 s?

I put 2.1 not sure if it's right.


2.) An airplane starts from rest and accelerates at 10.8 m/s to power of ((2)). What is its speed at the end of a 400m long runway.

I don't really know where to begin.

3.) A displacement vector is 34.0m in length and is directed 60.0 degrees east of north. What are the components of this vector?

Again I don't even know where to begin.

It would be nice if someone could explain these to me. Thanks in advance.

Alksjadf
 
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For question 1.) you are correct but remember your units.

For 2 you will need another of the kinematic equations which can be found here.

https://www.physicsforums.com/showpost.php?p=905663&postcount=2

Remember what information you're given and what you need, that will help you choose.

For 3.), think of the vecto as the hypotenuse of a right angled triangle the components are the lengths of the other two sides in the north and east directions. How would you find the lengths of two sides of a right angled triangle given the hypotenuse and an angle?
 
Oh ok so for number 3 i would have to use sin 60 = dy/34 and cos 60 = dx/34.

And for 2 I'm not sure what equation to use is the velocity and displacement the right one?

v^2 = v_0^2 + 2 a \Delta x
 
Your cos and sin will be the other way round because the angle is measured from the y-axis (north).

Yes you could use the equation you've stated for 2.
 
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