How Do You Calculate the Change in Velocity for a Van Turning a Corner?

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To calculate the change in velocity for a van turning a corner, it's essential to understand that the van initially travels north at 28 m/s and then turns to head 40° east of north at 25 m/s. The concept of "east of north" indicates a rotation of the direction, not a simple addition of speeds. To find the change in velocity, one must convert the velocities into Cartesian coordinates and then subtract the initial vector from the final vector, as the Pythagorean theorem does not apply due to the angle not being 90 degrees. Drawing a diagram to visualize the vectors and their components can aid in understanding the calculation of the change in velocity.
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hello everybody
can somebody help me how to do this exercise. i don't know how to start...
exercise:
a van is traveling north at a speed of 28m/s.after turning a corner it is heading 40° east of north at 25 m/s.
work out the change in velocity of the van.

first of all what i don't understand is what exactly do they mean with east of north? do they mean northeast?
i tried to work out the resultant but thaats not possible as i can't use pythagoras as there's no right angle.

i would be really happy to get some help
thanks in advance :)
 
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What does change mean? Surely not addition.

40 degrees east of north means that you "rotate" your direction for 40 degrees east of north. :)
 
Let's start off in polar coordinates assuming that north is in the direction of the positive y-axis and call the first velocity/angle V1 and the second V2.
(in polar coordinates, an ordered pair is (r,theta) where r is the magnitude and theta is the angle)

V1 = (28, 90)
V2 = (25, 50)

Now if you're looking for a simple change in velocity, it'd be 3m/s. But I'm guessing that'd be too easy. If you convert the polar coordinates to cartesian, then you can find the change in velocity with respect to y (north) and x (east).
 
noon0788 said:
Let's start off in polar coordinates assuming that north is in the direction of the positive y-axis and call the first velocity/angle V1 and the second V2.
(in polar coordinates, an ordered pair is (r,theta) where r is the magnitude and theta is the angle)

V1 = (28, 90)
V2 = (25, 50)

Now if you're looking for a simple change in velocity, it'd be 3m/s. But I'm guessing that'd be too easy. If you convert the polar coordinates to cartesian, then you can find the change in velocity with respect to y (north) and x (east).

thanks for the explanation
but i still don't understand it.
whats a cartesian?
in class we used to draw the vertical and horizontal component for such exercises.
but i don't know how to apply it here
 
radou said:
What does change mean? Surely not addition.

40 degrees east of north means that you "rotate" your direction for 40 degrees east of north. :)

i thought the change is 3 but that is apparently wrong:rolleyes:
 
koat said:
thanks for the explanation
but i still don't understand it.
whats a cartesian?
in class we used to draw the vertical and horizontal component for such exercises.
but i don't know how to apply it here

Draw a map with north on the vertical axis.
The original velocity is 28 m/s in the vertical direction on the map.
The speed after the change is 25 m/s, but the direction has also shifted to the right, forming a 40 degree angle with the original vector at the origin.

What is the difference between those two vectors?
 
PhaseShifter said:
Draw a map with north on the vertical axis.
The original velocity is 28 m/s in the vertical direction on the map.
The speed after the change is 25 m/s, but the direction has also shifted to the right, forming a 40 degree angle with the original vector at the origin.

What is the difference between those two vectors?

is that right?
28²+25² and then square root it?
 
koat said:
is that right?
28²+25² and then square root it?

Not at all.
1) The vectors form a 40 degree angle, not a 90 degree angle The Pythagorean theorem is useless here.
2) The change is the difference between the vectors, not the sum--so even if the angle had been 90 degrees the Pythagorean theorem would be useless.
3) The problem asks for change in velocity, not change in speed. The answer will be a vector, not a scalar.

You need to learn how to subtract vectors, and you need to learn how to find the vertical and horizontal components of a diagonal vector (this will involve some trigonometry).

Try drawing a diagram of the problem and see if that helps you understand better.
 
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